Cho biểu thức:
M = x + 2 x - 3 + x + 1 x - 2 - 3 . x - 1 x - 5 x + 6 với x ≥ 0 , x ≠ 4 , x ≠ 9
a) Thu gọn biểu thức M.
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14 tháng 3 2022
a. \(A=\left(\dfrac{2-3x}{x^2+2x-3}-\dfrac{x+3}{1-x}-\dfrac{x+1}{x+3}\right):\dfrac{3x+12}{x^3-1}\left(ĐKXĐ:x\ne1;x\ne-3\right)\)
\(=\left(\dfrac{2-3x}{\left(x-1\right)\left(x+3\right)}+\dfrac{x+3}{x-1}-\dfrac{x+1}{x+3}\right):\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\left(\dfrac{2-3x}{\left(x-1\right)\left(x+3\right)}+\dfrac{\left(x+3\right)^2}{\left(x-1\right)\left(x+3\right)}-\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+3\right)}\right):\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{2-3x+x^2+6x+9-x^2+1}{\left(x-1\right)\left(x+3\right)}:\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{3x+12}{\left(x-1\right)\left(x+3\right)}:\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{3x+12}{\left(x-1\right)\left(x+3\right)}.\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{3x+12}=\dfrac{x^2+x+1}{x+3}\)
\(M=A.B=\dfrac{x^2+x+1}{x+3}.\dfrac{x^2+x-2}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^2+x-2}{x+3}\)
b. -Để M thuộc Z thì:
\(\left(x^2+x-2\right)⋮\left(x+3\right)\)
\(\Rightarrow\left(x^2+3x-2x-6+4\right)⋮\left(x+3\right)\)
\(\Rightarrow\left[x\left(x+3\right)-2\left(x+3\right)+4\right]⋮\left(x+3\right)\)
\(\Rightarrow4⋮\left(x+3\right)\)
\(\Rightarrow x+3\in\left\{1;2;4;-1;-2;-4\right\}\)
\(\Rightarrow x\in\left\{-2;-1;1;-4;-5;-7\right\}\)
c. \(A^{-1}-B=\dfrac{x+3}{x^2+x+1}-\dfrac{x^2+x-2}{x^3-1}\)
\(=\dfrac{x+3}{x^2+x+1}-\dfrac{x^2+x-2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{\left(x+3\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{x^2+x-2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^2-x+3x-3-x^2-x+2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)
\(=\dfrac{1}{x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}\)
\(Max=\dfrac{4}{3}\Leftrightarrow x=\dfrac{-1}{2}\)
PN
9 tháng 5 2021
a, Do \(x=-3\)\(=>A=\frac{x+3}{x+2}=\frac{-3+3}{-3+2}=\frac{0}{-1}=0\)
Vậy A = 0 khi x = -3
b, Ta có : \(B=\frac{x}{x+1}+\frac{2}{x-1}-\frac{4}{x^2-1}=\frac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}+\frac{2\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{4}{x^2-1}\)
\(=\frac{x^2-x+2x-2}{x^2-1}=\frac{x\left(x-1\right)+2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=\frac{\left(x+2\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\frac{x+2}{x+1}\)(đpcm)
9 tháng 5 2021
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16 tháng 4 2021
Câu 2:
2) Ta có: \(M=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)
\(=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{2\sqrt{x}-9-x+9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{-\sqrt{x}}{\sqrt{x}-3}\)
MN
16 tháng 4 2021
Câu 2 :
Gọi : vận tốc của người đi chậm là : x (km/h) ( x > 0 )
Vận tốc của người đi nhanh : x + 4 (km/h)
Vi : người đi chậm đến muộn hơn : 45 phút \(=\dfrac{3}{4}\left(h\right)\)
Khi đó :
\(\dfrac{36}{x}-\dfrac{36}{x+4}=\dfrac{3}{4}\)
\(\Leftrightarrow\left[36\cdot\left(x+4\right)-36x\right]\cdot4=3x\cdot\left(x+4\right)\)
\(\Leftrightarrow3x^2+12x-144=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=12\left(n\right)\\x=16\left(l\right)\end{matrix}\right.\)
24 tháng 12 2017
Bài 1.
a) ( x - 2)2 - ( x + 3)( x - 3)= 17
=> x2 - 4x + 4 - x2 + 9 - 17 = 0
=> -4x - 4 = 0
=> -4( x + 1 ) = 0
=> x = -1
Vậy,...
b)4( x - 3)2 - ( 2x - 1)( 2x + 1) = 10
=> 4( x2 - 6x + 9) - 4x2 + 1 - 10 = 0
=> - 24x + 36 - 9 = 0
=> -24x + 27 = 0
=> -3( 8x - 9) = 0
=> x = \(\dfrac{9}{8}\)
Vậy,...
c) ( x - 4)2 - ( x - 2)( x + 2)= 36
=> x2 - 8x + 16 - x2 + 4 - 36 = 0
=> -8x - 16 = 0
=> -8( x + 2) = 0
=> x = -2
d) ( 2x + 3)2 - ( 2x + 1)( 2x - 1) = 10
=> 4x2 + 12x + 9 - 4x2 + 1 - 10 = 0
=> 12x = 0
=> x = 0
Vậy,...
Bài 2.
\(\dfrac{3x^2+3x}{\left(x+1\right)\left(2x-6\right)}\)
a) ĐKXĐ : ( x + 1)( 2x - 6) # 0
=> 2( x + 1)( x - 3) # 0
=> x # -1 ; x # 3
Vậy,...
b) Để P = 1
=> \(\dfrac{3x^2+3x}{\left(x+1\right)\left(2x-6\right)}=1\)
=> \(\dfrac{3x\left(x+1\right)}{2\left(x+1\right)\left(x-3\right)}=\dfrac{3x}{2\left(x-3\right)}=1\)
=> 3x = 2x - 6
=> x = -6 ( thỏa mãn ĐKXĐ)
Vậy,...
Bài 3.
P = \(\dfrac{x}{x-1}+\dfrac{x^2+1}{1-x^2}\)
a) Để P có nghĩa tức P xác định .
ĐKXĐ : x - 1 # 0 => x # 1
* 1 - x2 # 0 => x # 1 ; x # -1
Vậy,...
b) P = \(\dfrac{x}{x-1}+\dfrac{x^2+1}{1-x^2}\)
P = \(\dfrac{x^2+x-x^2-1}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x+1}\)( x# 1; x# -1)
c) Để P = -1 thì :
\(\dfrac{1}{x+1}=-1\)
=> -x - 1 = 1
=> x = -2 ( thỏa mãn ĐKXĐ )
Vậy,...
MN
1
TM
22 tháng 8 2021
a/ \(M=\left(1+\dfrac{x^2}{x^2+1}\right):\left(\dfrac{1}{x-1}-\dfrac{2x}{x^3+x-x^2-1}\right)\)
\(=\dfrac{2x^2+1}{x^2+1}:\dfrac{x-1}{x^2+1}\)
\(=\dfrac{\left(2x^2+1\right)\left(x^2+1\right)}{\left(x^2+1\right)\left(x-1\right)}\)
\(=\dfrac{2x^2+1}{x-1}\)
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b/ Thay x=-4 vào M ta được:
\(\dfrac{2.\left(-4\right)^2+1}{-4-1}=-\dfrac{33}{5}\)
Vậy: Giá trị của M tại x=-4 là \(-\dfrac{33}{5}\)
11 tháng 7 2021
Ta có: M=A+B
\(=\dfrac{x-\sqrt[3]{x}}{x-1}+\dfrac{1}{\sqrt[3]{x}-1}+\dfrac{1}{\sqrt[3]{x^2}+\sqrt[3]{x}+1}\)
\(=\dfrac{x-\sqrt[3]{x}}{\left(\sqrt[3]{x}-1\right)\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}+\dfrac{\sqrt[3]{x^2}+\sqrt[3]{x}+1+\sqrt[3]{x}-1}{\left(\sqrt[3]{x}-1\right)\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}\)
\(=\dfrac{x+\sqrt[3]{x}+\sqrt[3]{x^2}}{\left(\sqrt[3]{x}-1\right)\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}\)
\(=\dfrac{\sqrt[3]{x}}{\sqrt[3]{x}-1}\)
a)