Tìm số hữu tỉ biết
a) (x-1)\(^5\) = - 243
b) \(\dfrac{x+2}{11}+\dfrac{x+2}{12}+\dfrac{x+2}{13}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)
c)x - 2\(\sqrt{x}\)= 0 (x \(\ge\)0)
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Những câu hỏi liên quan
7 tháng 3 2017
a) Ta có:
(x - 1)5 = - 243
=> (x - 1)5 = (-3)5
=> x - 1 = - 3
=> x = -3 + 1
=> x = -2
Vậy x = -2
b) Ta có:
\(\dfrac{x+2}{11}+\dfrac{x+2}{12}+\dfrac{x+2}{13}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)
\(\Rightarrow\left(x+2\right).\dfrac{1}{11}+\left(x+2\right).\dfrac{1}{12}+\left(x+2\right).\dfrac{1}{13}=\left(x+2\right).\dfrac{1}{14}+\left(x+2\right).\dfrac{1}{15}\)
=> \(\left(x+2\right).\left(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}\right)=\left(x+2\right).\left(\dfrac{1}{14}+\dfrac{1}{15}\right)\)
=> \(\left(x+2\right).\dfrac{431}{1716}=\left(x+2\right).\dfrac{29}{210}\)
=> \(\left(x+2\right).\dfrac{431}{1716}-\left(x+2\right).\dfrac{29}{210}=0\)
=> (x + 2).(\(\dfrac{431}{1716}-\dfrac{29}{210}\)) = 0
mà \(\dfrac{431}{1716}-\dfrac{29}{210}\) \(\ne\) 0
=> x + 2 = 0
=> x = -2
Vậy x = -2
c) Ta có :
\(\left|3x-2\right|+5x=4x-10\)
=> \(\left|3x-2\right|=4x-5x-10\)
=> \(\left|3x-2\right|=-x-10\)
=> 3x - 2 = -x - 10
hoặc 3x - 2 = -(-x -10)
*) Nếu 3x - 2 = -x - 10
=> 3x + x = -10 + 2
=> 4x = -8
=> x = -2
*) Nếu 3x - 2 = -(-x -10)
=> 3x - 2 = x +10
=> 3x - x = 10 + 2
=> 2x = 12
=> x = 6
Vậy x = -2 hoặc x = 6
AN
25 tháng 12 2017
a, \(\left(x-1\right)^5=-243\)
=> \(\left(x-1\right)^5=\left(-3\right)^5\)
=> x-1= -3
=> x= -2
AN
25 tháng 12 2017
b, \(\dfrac{x+2}{11}+\dfrac{2+x}{12}+\dfrac{x+2}{13}=\dfrac{2+x}{14}+\dfrac{x+2}{15}\)
=> \(\dfrac{x+2}{11}+\dfrac{2+x}{12}+\dfrac{x+2}{13}-\dfrac{2+x}{14}+\dfrac{x+2}{15}=0\)
=>\(\dfrac{x+2+2+x+x+2-2+x+x+2}{11+12+13-14+15}\)
=> \(\dfrac{x+2}{37}=0\)
=> x+2= 0
=> x=-2
G
10 tháng 6 2017
Có:
\(\dfrac{x+2}{11}+\dfrac{x+2}{12}+\dfrac{x+2}{13}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)
\(\Leftrightarrow\dfrac{x+2}{11}+\dfrac{x+2}{12}+\dfrac{x+2}{13}-\dfrac{x+2}{14}-\dfrac{x+2}{15}=0\)
\(\Leftrightarrow\left(x+2\right)\left(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15}\right)=0\)
Dấu "=" xảy ra:
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15}=0\end{matrix}\right.\)
Vì \(\left(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15}\right)\ne0\)
\(\Leftrightarrow x-2=0\)
\(\Rightarrow x=0+2=2\)
Vậy \(x=2\).
Học tốt!
S
11 tháng 6 2017
\(\dfrac{x+2}{11}+\dfrac{x+2}{12}+\dfrac{x+2}{13}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)
\(\Rightarrow\left(\dfrac{1}{11}+\dfrac{1}{12}\right)\left(x+2\right)+\dfrac{x+2}{13}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)
\(\Rightarrow\dfrac{23\left(x+2\right)}{132}+\dfrac{x+2}{13}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)
\(\Rightarrow\left(\dfrac{23}{132}+\dfrac{1}{13}\right)\left(x+2\right)=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)
\(\Rightarrow\dfrac{431\left(x+2\right)}{1716}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)
\(\Rightarrow\dfrac{431\left(x+2\right)}{1716}=\left(\dfrac{1}{14}+\dfrac{1}{15}\right)\left(x+2\right)\)
\(\Rightarrow\dfrac{431\left(x+2\right)}{1716}=\dfrac{29\left(x+2\right)}{210}\)
\(\Rightarrow\dfrac{431\left(x+2\right)}{1716}-\dfrac{29\left(x+2\right)}{210}=0\)
\(\Rightarrow\left(\dfrac{431}{6.286}-\dfrac{29}{6.35}\right)\left(x+2\right)=0\)
\(\Rightarrow\dfrac{1}{6}\left(\dfrac{431}{286}-\dfrac{29}{35}\right)\left(x+2\right)=-2\)
17 tháng 4 2021
a) Ta có: \(\dfrac{x-2}{15}+\dfrac{x-3}{14}+\dfrac{x-4}{13}+\dfrac{x-5}{12}=4\)
\(\Leftrightarrow\dfrac{x-2}{15}-1+\dfrac{x-3}{14}-1+\dfrac{x-4}{13}-1+\dfrac{x-5}{12}-1=0\)
\(\Leftrightarrow\dfrac{x-17}{15}+\dfrac{x-17}{14}+\dfrac{x-17}{13}+\dfrac{x-17}{12}=0\)
\(\Leftrightarrow\left(x-17\right)\left(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}\right)=0\)
mà \(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}>0\)
nên x-17=0
hay x=17
Vậy: x=17
17 tháng 4 2021
b) Ta có: \(\dfrac{x+1}{19}+\dfrac{x+2}{18}+\dfrac{x+3}{17}+...+\dfrac{x+18}{2}+18=0\)
\(\Leftrightarrow\dfrac{x+1}{19}+1+\dfrac{x+2}{18}+1+\dfrac{x+3}{17}+1+...+\dfrac{x+18}{2}+1=0\)
\(\Leftrightarrow\dfrac{x+20}{19}+\dfrac{x+20}{18}+\dfrac{x+20}{17}+...+\dfrac{x+20}{2}=0\)
\(\Leftrightarrow\left(x+20\right)\left(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}\right)=0\)
mà \(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}>0\)
nên x+20=0
hay x=-20
Vậy: x=-20
21 tháng 8 2017
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Rightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
\(\Rightarrow x+1=0\Rightarrow x=-1\)
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Rightarrow x+2004=0\Rightarrow x=-2004\)
21 tháng 8 2017
a, \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Rightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
Do \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\)
\(\Rightarrow x+1=0\Rightarrow x=-1\)
Vậy x = -1
b, \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
Vì \(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\)
\(\Rightarrow x+2004=0\Rightarrow x=-2004\)
Vậy...
14 tháng 10 2021
\(a,A=4\sqrt{3}-5\sqrt{3}+2-\sqrt{3}=2-2\sqrt{3}\\ B=\dfrac{x+2\sqrt{x}+8+2\sqrt{x}-8}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-4}\\ b,B-\dfrac{1}{2}A=\dfrac{\sqrt{x}}{\sqrt{x}-4}-\dfrac{1}{2}\left(2-2\sqrt{3}\right)=0\\ \Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-4}=1+\sqrt{3}\\ \Leftrightarrow\sqrt{x}=\left(1+\sqrt{3}\right)\left(\sqrt{x}-4\right)\Leftrightarrow\sqrt{x}=\sqrt{x}-4\sqrt{3}+\sqrt{3x}-4\\ \Leftrightarrow\sqrt{3x}=4\sqrt{3}+4\\ \Leftrightarrow\sqrt{x}=\dfrac{4\sqrt{3}+4}{\sqrt{3}}\\ \Leftrightarrow\sqrt{x}=\dfrac{12+4\sqrt{3}}{3}\\ \Leftrightarrow x=\dfrac{192+96\sqrt{3}}{9}=\dfrac{64+32\sqrt{3}}{3}\)
12 tháng 9 2021
\(a,\dfrac{-5}{x-3}< 0\Leftrightarrow x-3>0\left(-5< 0\right)\Leftrightarrow x>3\\ b,\dfrac{3-x}{x^2+1}\ge0\Leftrightarrow3-x\ge0\left(x^2+1>0\right)\Leftrightarrow x\le3\\ c,\dfrac{\left(x-1\right)^2}{x-2}< 0\Leftrightarrow x-2< 0\left[\left(x-1\right)^2\ge0\right]\Leftrightarrow x< 2\)
PT
19 tháng 8 2017
a) Ta có : \(x - 2xy + y - 3 = 0\)
\(\Rightarrow-2xy+x+y=3\)
\(\Rightarrow-2.\left(-2xy+x+y\right)=-2.3\)
\(\Rightarrow4xy-2x-2y=-6\)
\(\Rightarrow4xy-2x-2y+1=-6+1\)
\(\Rightarrow2x.\left(2y-1\right).\left(2y-1\right)=-5\)
\(\Rightarrow\left(2y-1\right).\left(2x-1\right)=-5=1.\left(-5\right)=-5.1=\left(-1\right).5=5.\left(-1\right)\)
Tự lập bảng đi -.-
26 tháng 3 2018
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Nhân từng vế bất đẳng thức ta được : (xyz)2 = 36xyz + Nếu một trong các số x,y,z bằng 0 thì 2 số còn lại cũng bằng 0 + Nếu cả 3 số x,y,z khác 0 thì chia 2 vế cho xyz ta được xyz = 36 + Từ xyz =36 và xy = z ta được z2 = 36 nên z = 6; z = -6 + Từ xyz =36 và yz = 4x ta được 4x2 = 36 nên x = 3; x = -3 + Từ xyz =36 và - Nếu z = 6 thì x và y cùng dấu nên x = 3, y = 2 hoặc x = -3 , y = -2 - Nếu z = -6 thì x và y trái dấu nên x = 3 ; y = -2 hoặc x = -3; y=2 |
Vậy có 5 bộ số (x, y, z) thoã mãn: (0,0,0); (3,2,6);(-3,-2,6);(3,-2,-6);(-3,2.-6)
a, \(\left(x-1\right)^5=-243\)
\(\Leftrightarrow\left(x-1\right)^5=-3^5\)
\(\Leftrightarrow x-1=-3\Leftrightarrow x=-2\)
b,\(\dfrac{x+2}{11}+\dfrac{x+2}{12}+\dfrac{x+2}{13}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)
\(\dfrac{x+2}{11}+\dfrac{x+2}{12}+\dfrac{x+2}{13}-\dfrac{x+2}{14}-\dfrac{x+2}{15}=0\)
\(\Leftrightarrow\left(x+2\right).\left(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15}\right)=0\)
\(do\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15}\ne0\)
\(\Rightarrow x+2=0\Leftrightarrow x=-2\)
c, \(x-2\sqrt{x}=0\Leftrightarrow\sqrt{x^2}-2\sqrt{x}=0\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\\sqrt{x}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=\sqrt{2}\end{matrix}\right.\)