a: Ta có: \(4x\left(x-7\right)-4x^2=56\)

\(\Leftrightarrow4x^2-7x-4x^2=56\)

hay x=-8

b: Ta có: \(12x\left(3x-2\right)-\left(4-6x\right)=0\)

\(\Leftrightarrow36x^2-24x-4+6x=0\)

\(\Leftrightarrow36x^2-18x-4=0\)

\(\text{Δ}=\left(-18\right)^2-4\cdot36\cdot\left(-4\right)=900\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{18-30}{72}=\dfrac{-1}{6}\\x_2=\dfrac{18+30}{72}=\dfrac{2}{3}\end{matrix}\right.\)

c: Ta có: \(4\left(x-5\right)-\left(x-5\right)^2=0\)

\(\Leftrightarrow\left(x-5\right)\left(4-x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=9\end{matrix}\right.\)

a: Ta có: \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)

\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{23}{7}\end{matrix}\right.\)

c: Ta có: \(\left(x-3\right)^2-4=0\)

\(\Leftrightarrow\left(x-5\right)\cdot\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)

b. 

PT $\Leftrightarrow (5x^2-2x+10)^2-(3x^2+10x-8)^2=0$

$\Leftrightarrow (5x^2-2x+10-3x^2-10x+8)(5x^2-2x+10+3x^2+10x-8)=0$

$\Leftrightarrow (2x^2-12x+18)(8x^2+8x+2)=0$

$\Leftrightarrow (x^2-6x+9)(4x^2+4x+1)=0$

$\Leftrightarrow (x-3)^2(2x+1)^2=0$

$\Leftrightarrow (x-3)(2x+1)=0$

$\Leftrightarrow x-3=0$ hoặc $2x+1=0$

$\Leftrightarrow x=3$ hoặc $x=-\frac{1}{2}$

d.

$x^2-2x=24$

$\Leftrightarrow x^2-2x-24=0$

$\Leftrightarrow (x+4)(x-6)=0$
$\Leftrightarrow x+4=0$ hoặc $x-6=0$

$\Leftrightarrow x=-4$ hoặc $x=6$

\(a,\dfrac{3}{4}x-\dfrac{2}{3}x=\dfrac{2}{7}\times\dfrac{1}{6}+\dfrac{5}{7}\times\dfrac{1}{6}\)

\(\dfrac{9}{12}x-\dfrac{8}{12}x=\dfrac{1}{6}\times\left(\dfrac{2}{7}+\dfrac{5}{7}\right)\)

\(\dfrac{1}{12}x=\dfrac{1}{6}\)

\(x=\dfrac{1}{6}:\dfrac{1}{12}\)

\(x=\dfrac{1}{6}\times12\)

\(x=2\)

a) \(\Rightarrow72-20x-36x+84=30x-240-6x-84\)

\(\Rightarrow80x=480\Rightarrow x=6\)

b) \(\Rightarrow15x+25-8x+12=5x+6x+36+1\)

\(\Rightarrow4x=0\Rightarrow x=0\)

c) \(\Rightarrow10x-16-12x+15=12x-16+11\)

\(\Rightarrow14x=4\Rightarrow x=\dfrac{2}{7}\)

a) |2x - 5| + |3x + 1| = 6

Ta có: (5 - 2x) + (-3x - 1) = 6

=> 4 - 5x = 6

=> 5x = 4 - 6 = -2

\(\Rightarrow x=\frac{-2}{5}\), thỏa mãn \(x< \frac{-1}{3}\)

Ta có: (5 - 2x) + (3x + 1) = 6

=> 6 + x = 6

=> x = 6 - 6 = 0, thỏa mãn \(\frac{-1}{3}\le x< \frac{5}{2}\)

Ta có: (2x - 5) + (3x + 1) = 6

=> 5x - 4 = 6

=> 5x = 6 + 4 = 10

=> x = 10 : 5 = 2, không thỏa mãn \(x\ge\frac{5}{2}\)

Vậy \(\left[\begin{array}{nghiempt}x=\frac{-2}{5}\\x=0\end{array}\right.\) thỏa mãn đề bài

mình k có time

2x-5+3x+1=6

suy ra: x(2+3)+(-5+1) =6

suy ra: x5 + (-4) =6

suy ra: x5 = 10

suy ra: x=2

x-4+x-6=2

suy ra: 2x + ( -4-6) =2

suy ra:2x+(-10)=2

suy ra:2x = 12

suy ra: x=6

a, 2,8

b, -4

\(a,0.25x\) - \(\frac{2}{3}x\) =\(1\frac{1}{6}\)

\(\Rightarrow x\cdot\left(0.25-\frac{2}{3}\right)=1\frac{1}{6}\)

\(\Rightarrow x\cdot\frac{-5}{12}=\frac{7}{6}\)

\(\Rightarrow x=\frac{7}{6}:\frac{-5}{12}\)

\(\Rightarrow x=\frac{-14}{5}=-2.8\)

Lời giải:

Đặt 

Khi đó:

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