Question

1:tìm x

a; \(3x+\left|x-2\right|=8\)

b; \(5-\left|x-1\right|=4\)

2:tìm x

\(5\cdot\left(x-2\right)-4\cdot\left(1-3x\right)=\left|3-7\right|+2\cdot\left(1+2x\right)\)

3: tìm x

\(\left(x-2\right)\cdot\left(2x+1\right)-3\cdot\left(x+2\right)=4-5\cdot\left(1-x\right)\)

4:tìm x

\(1\dfrac{1}{2}\cdot\left(x-2\right)-\dfrac{x-5}{3}=3\dfrac{1}{3}\cdot\left(1-2x\right)-\dfrac{5\cdot\left(x+1\right)}{6}\)

5: tìm x

\(\left(x-3\right)\cdot\left(1-x\right)+\left(x-2\right)^2=\left(1-x\right)^2-2\cdot\left(1+x\right)\)

6: tìm x

\(\left(2x-1\right)^2-3\cdot\left(x+2\right)^2=4\cdot\left(x-2\right)-5\cdot\left(x-1\right)^2\)

Answer 1

1. a, 3x + |x - 2| = 8
<=> |x - 2| = 8 - 3x
Xét 2 TH :
TH1: x - 2 = 8 - 3x
<=> x + 3x = 8 + 2
<=> 4x = 10
<=> x = \(\dfrac{5}{2}\) (thỏa mãn)
TH2: x - 2 = -(8 - 3x)
<=> x - 2 = -8 + 3x
<=> -2 + 8 = 3x - x
<=> 6 = 2x
<=> x = 3 (thỏa mãn)
b, 5 - |x - 1| = 4
<=> |x - 1| = 1
<=> \(\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\) (thỏa mãn)
@Nguyễn Hoàng Vũ

Answer 2

2. 5.(x - 2) - 4.(1 - 3x) = |3 - 7| + 2.(1 + 2x)
<=> 5x - 10 - 4 + 12x = 4 + 2 + 4x
<=> 17x - 14 = 6 + 4x
<=> 17x - 4x = 6 + 14
<=> 13x = 20
<=> x = \(\dfrac{20}{13}\) (thỏa mãn)
@Nguyễn Hoàng Vũ

Answer 3

4. 1\(\dfrac{1}{2}\).(x - 2) - \(\dfrac{x-5}{3}\) = 3\(\dfrac{1}{3}\).(1 - 2x) - \(\dfrac{5.\left(x+1\right)}{6}\)

<=> \(\dfrac{3}{2}\).(x - 2) - \(\dfrac{x-5}{3}\) = \(\dfrac{10}{3}\).(1 - 2x) - \(\dfrac{5x+5}{6}\)

<=> \(\dfrac{3}{2}x-3-\dfrac{x}{3}+\dfrac{5}{3}=\dfrac{10}{3}-\dfrac{20}{3}x-\dfrac{5x}{6}-\dfrac{5}{6}\)

<=> \(\dfrac{3}{2}x-\dfrac{x}{3}+\dfrac{20}{3}x-\dfrac{5x}{6}=\dfrac{10}{3}-\dfrac{5}{6}-3+\dfrac{5}{3}\)

<=> 7x = \(\dfrac{7}{6}\)

<=> x = \(\dfrac{1}{6}\)
@Nguyễn Hoàng Vũ

Answer 4

3. (x - 2)(2x + 1) - 3(x + 2) = 4 - 5(1 - x)
<=> 2x(x - 2) + (x - 2) = 4 - 5 + 5x
<=> 2x² - 4 + x - 2 = -1 + 5x
<=> -4 - 2 + 1 = 5x - 2x² - x
<=> -5 = x(5 - 2x - 1)
<=> -5 = x(4 - 2x)
<=> -5 = 4x - 2x
<=> -5 = 2x
<=> x = \(-\dfrac{5}{2}\) (thỏa mãn)
@Nguyễn Hoàng Vũ

Answer 5

5. (x - 3)(1 - x) + (x - 2)² = (1 - x)² - 2(1 + x)
<=> (x - 3) - x(x - 3) + (x - 2)x + 2(x - 2) = (1 - x) - x(1 - x) - 2 - 2x
<=> x - 3 - x² + 3x + x² - 2x + 2x - 4 = 1 - x - x + x² - 2 - 2x
<=> x - x² + 3x + x² - 2x + 2x + x + x - x² + 2x = 1 - 2 + 3 + 4
<=> 7x = 6
<=> x = \(\dfrac{6}{7}\) (thỏa mãn)
@Nguyễn Hoàng Vũ

Answer 6

6. (2x - 1)² - 3(x + 2)² = 4(x - 2) - 5(x - 1)²
<=> (2x)² - 1 - 3x² - 6² = 4x - 8 - 5x² + 5²
<=> 4x² - 1 - 3x² - 36 = 4x - 8 - 5x² + 25
<=> 4x² - 3x² - 4x + 5x² = -8 - 25 + 1 + 36
<=> 2x = 4
<=> x = 2 (thỏa mãn)
@Nguyễn Hoàng Vũ

Answer 7

Có 1 học sinh gương mẫu nào đó của thầy Phú lại không tự làm bài tập mà cứ đăng bài dài dài,thế mà còn bày đặt nói người ta ~~~~~Ai kia cũng vậy chứ khác gì nhau đâu