a) \(\left|x\left(x-7\right)\right|=x\)

\(\Rightarrow\orbr{\begin{cases}x\left(x-7\right)=x\\x\left(x-7\right)=-x\end{cases}\Leftrightarrow\orbr{\begin{cases}x-7=1\\x-7=-1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=8\\x=6\end{cases}}}\)

b) \(\left|x-1,1\right|+\left|x+1,2\right|+\left|x+1,3\right|+\left|x+1,4\right|=5x\)

\(\Rightarrow x-1,1+x+1,2+x+1,3+x+1,4=5x\)

\(\Leftrightarrow4x+2,8=5x\)

\(\Leftrightarrow x=2,8\)

\(a.\)\(\left|x.\left(x-7\right)\right|=x\)( Đk: \(x\ge0\))

\(\Leftrightarrow\orbr{\begin{cases}x.\left(x-7\right)=x\\x.\left(x-7\right)=-x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x-7=x:x\\x-7=-x:x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x-7=1\\x-7=-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1+7\\x=-1+7\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=8\\x=6\end{cases}}\)

\(b.\)\(\left|x-1,1\right|+\left|x+1,2\right|+\left|x+1,3\right|+\left|x+1,4\right|=5x\)( Đk: \(5x\ge0\Leftrightarrow x\ge0\))

\(\Rightarrow x-1,1+x+1,2+x+1,3+x+1,4=5x\)

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

\(\Leftrightarrow4x+2,8=5x\)

\(\Leftrightarrow2,8=5x-4x\)

\(\Leftrightarrow x=2,8\)

\(c.\)\(7^{x+2}+2.7^{x-1}=345\)

\(\Leftrightarrow7^{x-1}.7^{x+3}+2.7^{x-1}=345\)

\(\Leftrightarrow7^{x-1}.\left(7^{x+3}+2\right)=345\)

            \(......................\)

Đến đây mk ko bt làm nữa, tự lm nhé !

len google di ban

mk chua hoc bai nay

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

\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)

\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x+10=3x^2-12x+20\)

\(\Leftrightarrow3x^2-7x-2=3x^2-12x+20\)

\(\Leftrightarrow-7x+12x=20+2\)

\(\Leftrightarrow5x=22\)

\(\Rightarrow x=\dfrac{22}{5}\)

tick cho mk nha

b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)

\(\Leftrightarrow24x^2+16x-9x-6-4x^2-23x-28=10x^2+3x-1\)

\(\Leftrightarrow20x^2-16x-34-10x^2-3x+1=0\)

\(\Leftrightarrow10x^2-19x-33=0\)

\(\Delta=\left(-19\right)^2-4.10.\left(-33\right)=1320\)

\(x_1=3;x_2=\dfrac{-11}{10}\)

Tick cho mk nha

c) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)=4\)

\(\Leftrightarrow21x-15x^2-35+25x-4x+15x^2-4=4\)

\(\Leftrightarrow42x-39=4\)

\(\Leftrightarrow42x=4+39\)

\(\Leftrightarrow42x=43\)

\(\Rightarrow x=\dfrac{43}{42}\)

Tick cho mk nha

\(a,\left(3x+4\right)\left(3x-4\right)-\left(2x+5\right)^2=\left(x-5\right)^2+\left(2x+1\right)^2-\left(x^2-2x\right)+\left(x-1\right)^2\\ \Leftrightarrow\left(9x^2-16\right)-\left(4x^2+20x+25\right)=x^2-10x+25+4x^2+4x+1-x^2+2x+x^2-2x+1\\ \Leftrightarrow9x^2-16-4x^2-20x-25=5x^2-6x+27\\ \Leftrightarrow5x^2-20x-41=5x^2-5x+27\\ \Leftrightarrow-15x=68\\ \Leftrightarrow x=-\dfrac{68}{15}\)Vậy..

Câu sau cũng tương tự nhé

\(8,1-\left(x-6\right)=4\left(2-2x\right)\)

\(\Leftrightarrow1-x+6=8-8x\)

\(\Leftrightarrow-x+8x=8-1-6\)

\(\Leftrightarrow7x=1\)

\(\Leftrightarrow x=\dfrac{1}{7}\)

\(9,\left(3x-2\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)

\(10,\left(x+3\right)\left(x^2+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\varnothing\end{matrix}\right.\)

`8)1-(x-5)=4(2-2x)`

`<=>1-x+5=8-6x`

`<=>5x=2<=>x=2/5`

`9)(3x-2)(x+5)=0`

`<=>[(x=2/3),(x=-5):}`

`10)(x+3)(x^2+2)=0`

  Mà `x^2+2 > 0 AA x`

 `=>x+3=0`

`<=>x=-3`

`11)(5x-1)(x^2-9)=0`

`<=>(5x-1)(x-3)(x+3)=0`

`<=>[(x=1/5),(x=3),(x=-3):}`

`12)x(x-3)+3(x-3)=0`

`<=>(x-3)(x+3)=0`

`<=>[(x=3),(x=-3):}`

`13)x(x-5)-4x+20=0`

`<=>x(x-5)-4(x-5)=0`

`<=>(x-5)(x-4)=0`

`<=>[(x=5),(x=4):}`

`14)x^2+4x-5=0`

`<=>x^2+5x-x-5=0`

`<=>(x+5)(x-1)=0`

`<=>[(x=-5),(x=1):}`

\(11,=>\left[{}\begin{matrix}5x-1=0\\x^2-9=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=3\\x=-3\end{matrix}\right.\\ 12,=>\left(x+3\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+3=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\\ 13,=>x\left(x-5\right)-4\left(x-5\right)=0\\ =>\left(x-4\right)\left(x-5\right)=0\\ =>\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

\(14,=>x^2+5x-x-5=0\\ =>x\left(x+5\right)-\left(x+5\right)=0\\ =>\left(x-1\right)\left(x+5\right)=0\\ =>\left[{}\begin{matrix}x-1=0\\x+5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)

1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)

=>-13x=0

=>x=0

2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

=>3x=13

=>x=13/3

3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)

=>-2x^2=0

=>x=0

4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

=>-8x=6-14=-8

=>x=1

`1)2x(x-5)-(3x+2x^2)=0`

`<=>2x^2-10x-3x-2x^2=0`

`<=>-13x=0`

`<=>x=0`

___________________________________________________

`2)x(5-2x)+2x(x-1)=13`

`<=>5x-2x^2+2x^2-2x=13`

`<=>3x=13<=>x=13/3`

___________________________________________________

`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`

`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`

`<=>x=0`

___________________________________________________

`4)5x(x-1)-(x+2)(5x-7)=0`

`<=>5x^2-5x-5x^2+7x-10x+14=0`

`<=>-8x=-14`

`<=>x=7/4`

___________________________________________________

`5)6x^2-(2x-3)(3x+2)=1`

`<=>6x^2-6x^2-4x+9x+6=1`

`<=>5x=-5<=>x=-1`

___________________________________________________

`6)2x(1-x)+5=9-2x^2`

`<=>2x-2x^2+5=9-2x^2`

`<=>2x=4<=>x=2`

1, \(-4x\left(x-7\right)+4x\left(x^2-5\right)=28x^2-13\)

\(\Leftrightarrow-4x^2+28x+4x^3-20x=28x^2-13\)

\(\Leftrightarrow-32x^2+8x+4x^3-13=0\)( vô nghiệm )

2, \(\left(4x^2-5x\right)\left(3x+2\right)-7x\left(x+5\right)=\left(-4+x\right)\left(-2x+3\right)+12x^3+2x^2\)

\(\Leftrightarrow12x^3-7x^2-10x-7x^2-35x=-2x^2+11x-12+12x^3+2x^2\)

\(\Leftrightarrow12x^3-14x^2-45x=11x-12+12x^3\)

\(\Leftrightarrow-14x^2-56x-12=0\)( vô nghiệm )

Mình làm riêng ra nhá , chứ nhiều quá nên thông cảm cho mình :))

1. \(-4x\left(x-7\right)+4x\left(x^2-5\right)=28x^2-13\)

=> \(-4x^2+28x+4x^3-20x=28x^2-13\)

=> \(-4x^2+4x^3+\left(28x-20x\right)=28x^2-13\)

=> \(-4x^2+4x^3+8x-28x^2+13=0\)

=> \(\left(-4x^2-28x^2\right)+4x^3+8x+13=0\)

=> \(-32x^2+4x^3+8x+13=0\)

=> vô nghiệm

2. \(\left(4x^2-5x\right)\left(3x+2\right)-7x\left(x+5\right)=\left(-4+x\right)\left(-2x+3\right)+12x^3+2x^2\)

=> \(4x^2\left(3x+2\right)-5x\left(3x+2\right)-7x\left(x+5\right)=-4\left(-2x+3\right)+x\left(-2x+3\right)+12x^3+2x^2\)

=> \(12x^3+8x^2-15x^2-10x-7x^2-35x=8x-12-2x^2+3x+12x^3+2x^2\)

=> \(12x^3+8x^2-15x^2-10x-7x^2-35x-8x+12+2x^2-3x-12x^3-2x^2=0\)

=> \(\left(12x^3-12x^3\right)+\left(8x^2-15x^2-7x^2+2x^2-2x^2\right)+\left(-10x-35x-8x-3x\right)+12=0\)

=> \(-14x^2-56x+12=0\)

=> .... tự tìm

Câu c dấu bằng chỗ nào ?

\(-4x\left(x-7\right)+4x\left(x^2-5\right)=28x^2-13\)

\(< =>-4x^2+28x+4x^3-20x=28x^2-13\)

\(< =>4x^3-\left(4x^2+28x^2\right)+8x+13=0\)

\(< =>4x^3-32x^2+8x+13=0\)

do cái này nghiệm màu mè nên mình sẽ làm cách khá khó hiểu 

\(< =>x^3-7x^2+2x+\frac{13}{4}=0\)

Đặt \(x=y+\frac{7}{3}\)khi đó phương trình trở thành 

\(\left(y+\frac{7}{3}\right)^3-7\left(y+\frac{7}{3}\right)^2+2\left(y+\frac{7}{3}\right)+\frac{13}{4}=0\)

\(< =>y^3+3y^2\frac{7}{3}+3y\frac{49}{9}-7\left(y^2+\frac{14y}{3}+\frac{49}{9}\right)+2y+\frac{14}{3}+\frac{13}{4}=0\)

\(< =>y^3+7y^2+\frac{49y}{3}-7y^2-\frac{98y}{3}-\frac{343}{9}+2y+\frac{95}{12}=0\)

\(< =>y^3-\frac{43y}{3}-\frac{1087}{36}=0\)

Đặt \(y=u+v\)sao cho \(uv=\frac{43}{9}\)khi đó pt trở thành 

\(\left(u+v\right)^3-\frac{43\left(u+v\right)}{3}-\frac{1087}{36}=0\)

\(< =>u^3+v^3+3uv\left(u+v\right)-\left(u+v\right).\frac{43}{3}-\frac{1087}{36}=0\)

\(< =>u^3+v^3+\left(u+v\right)\left(3uv-\frac{43}{3}\right)-\frac{1087}{36}=0\)

\(< =>u^3+v^3=\frac{1087}{36}\)(*) (do \(uv=\frac{43}{9}\Leftrightarrow3uv-\frac{43}{3}=0\))

Ta có \(uv=\frac{43}{9}\Leftrightarrow u^3v^3=\frac{79507}{729}\)(**)

Từ (*) và (*) Suy ra được \(\hept{\begin{cases}u^3+v^3=\frac{1087}{36}\\u^3v^3=\frac{79507}{729}\end{cases}}\)

đến đây dễ rồi nhé ^^

a) Ta có: \(\left(6x-2\right)^2+\left(5x-2\right)^2-4\left(3x-1\right)\left(5x-2\right)=0\)

\(\Leftrightarrow\left(6x-2\right)^2-2\cdot\left(6x-2\right)\left(5x-2\right)+\left(5x-2\right)^2=0\)

\(\Leftrightarrow\left(6x-2-5x+2\right)^2=0\)

\(\Leftrightarrow x^2=0\)

hay x=0

Vậy: x=0

b) Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)

\(\Leftrightarrow x^3-1-x\left(x^2-4\right)-5=0\)

\(\Leftrightarrow x^3-6-x^2+4x=0\)

\(\Leftrightarrow4x-6=0\)

\(\Leftrightarrow4x=6\)

hay \(x=\frac{3}{2}\)

Vậy: \(x=\frac{3}{2}\)

c) Ta có: \(\left(x-1\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+3\left(x^2-4\right)=2\)

\(\Leftrightarrow x^3-3x^2+3x-1-\left(x^3+27\right)+3x^2-12-2=0\)

\(\Leftrightarrow x^3+3x-15-x^3-27=0\)

\(\Leftrightarrow3x-42=0\)

\(\Leftrightarrow3x=42\)

hay x=14

Vậy: x=14