Tìm x, biết:
a ) 2 | x | + 1 2 = 2 b ) x + 1 2 − 1 6 = 1 c ) 1 4 − x +5 1 8 = 6 1 8 d ) 9 2 − x − 3 4 = − 1 2
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a,(x+2)(x+3)-(x+2)(x+5)=6
<=>(x+2)(x+3-x-5)=6
<=>(x+2).-2=6
<=>x+2=-3
<=>x=-3-2=-5
k rồi mình làm tiếp cho
1
\(\left(x-2\right):2.3=6\)
\(\Leftrightarrow\left(x-2\right):2=2\)
\(\Leftrightarrow\left(x-2\right)=4\)
\(\Leftrightarrow x=4+2=6\)
c) ta có
\(\left[\left(2x+1\right)+1\right]m:2=625\)
\(\Leftrightarrow\left[\left(2x+1\right)+1\right]\left\{\left[\left(2x+1\right)-1\right]:2+1\right\}=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-1:2+1=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-2+1=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-2=1249\)
\(\Leftrightarrow\left(2x+1\right)^2+1=1251\)
\(\Leftrightarrow\left(2x+1\right)^2=1250\)
...
2
\(\left(x-\frac{1}{2}\right).\frac{5}{3}=\frac{7}{4}-\frac{1}{2}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right).\frac{5}{3}=\frac{5}{4}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)=\frac{5}{4}:\frac{5}{3}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)=\frac{5}{4}.\frac{3}{5}\)
\(\Leftrightarrow x-\frac{1}{2}=\frac{3}{4}\)
\(\Leftrightarrow x=\frac{3}{4}+\frac{1}{2}=\frac{5}{4}\)
Bài 2:
a) Ta có: \(\left|x-2\right|=\left|4-x\right|\)
\(\Leftrightarrow x-2=4-x\)
\(\Leftrightarrow2x=6\)
hay x=3
b) Ta có: \(\left(\left|2x-1\right|-3\right)\cdot\left(-2\right)+\left(-5\right)=6\)
\(\Leftrightarrow\left(\left|2x-1\right|-3\right)\cdot\left(-2\right)=11\)
\(\Leftrightarrow\left|2x-1\right|-3=\dfrac{-11}{2}\)
\(\Leftrightarrow\left|2x-1\right|=\dfrac{-11}{2}+\dfrac{6}{2}=\dfrac{-5}{2}\)(Vô lý)
\(A=x^2-2x+10\)
\(A=\left(x^2-2x+1\right)+9\)
\(A=\left(x-1\right)^2+9\)
Mà \(\left(x-1\right)^2\ge0\)
\(\Rightarrow A\ge9\)
Dấu "=" xảy ra khi :
\(x-1=0\Leftrightarrow x=1\)
Vậy Min A = 9 khi x = 1
\(B=x^2-5x-7\)
\(B=\left(x^2-5x+\frac{25}{4}\right)-\frac{53}{4}\)
\(B=\left(x-\frac{5}{2}\right)^2-\frac{53}{4}\)
Mà \(\left(x-\frac{5}{2}\right)^2\ge0\)
\(\Rightarrow B\ge-\frac{53}{4}\)
Dấu "=" xảy ra khi :
\(x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\)
Vậy \(B_{Min}=-\frac{53}{4}\Leftrightarrow x=\frac{5}{2}\)
\(a,\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=0\\ \Rightarrow\left(x^3-27\right)+x\left(4-x^2\right)=0\\ \Rightarrow x^3-27+4x-x^3=0\\ \Rightarrow4x-27=0\\ \Rightarrow4x=27\\ \Rightarrow x=\dfrac{27}{4}\)
\(b,\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\\ \Rightarrow\left(x^3+3x^2+3x+1\right)-\left(x^3-3x^2+3x-1\right)-6\left(x^2-2x+1\right)=-10\\ \Rightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6+10=0\)
\(\Rightarrow12x+6=0\\ \Rightarrow12x=-6\\ \Rightarrow x=-\dfrac{1}{2}\)
Đáp án là B vì 12: -3 = -4; 12: -4 = -3; 12: -6 = -2;12: -12 = -1 và đáp ứng điều kiện a< -2
a ) x ∈ − 3 4 ; 3 4 b ) x ∈ 2 3 ; − 5 3 ; c ) x ∈ − 3 4 ; 5 4 d ) x ∈ 23 4 ; − 17 4