a) x^2+5x+6-x^2+7x-10-6=0

12x-10=0

12x=10

x=5/6

Cậu ơi giúp mình 2 câu dưới nữa ược không?

b)  3(2x - 1)(3x - 1) - (2x - 3)(9x - 1) = 0

<=>14x=0

<=>x=0

c) (3x + 2)(2x + 9) - (x + 2)(6x + 1) = (x + 1) - (x - 6)   

<=>2(9x+8)=7

<=>18x+16=7

<=>18x=-9

<=>x=-1/2

`x xx 6/7=5/14`

`=>x=5/14:6/7`

`=>x=5/14xx7/6`

`=>x=35/84`

`=>x=5/12`

Vậy `x=5/12`

__

`x:2/3=4/9`

`=>x=4/9xx2/3`

`=>x=8/27`

Vậy `x=8/27`

__

`x-1/4=3/2`

`=>x=3/2+1/4`

`=>x=6/4+1/4`

`=>x=7/4`

Vậy `x=7/4`

__

`x+4/5=8/9`

`=>x=8/9-4/5`

`=>x=40/45-36/45`

`=>x=4/45`

Vậy `x=4/45`

\(x\cdot\dfrac{6}{7}=\dfrac{5}{14}\)

\(x\)         \(=\dfrac{5}{14}:\dfrac{6}{7}\)

\(x\)           \(=\dfrac{5}{12}\)

\(x:\dfrac{2}{3}=\dfrac{4}{9}\)

\(x\)        \(=\dfrac{4}{9}\cdot\dfrac{2}{3}\)

\(x\)          \(=\dfrac{8}{27}\)

\(x-\dfrac{1}{4}=\dfrac{3}{2}\)

\(x\)          \(=\dfrac{3}{2}+\dfrac{1}{4}\)

\(x\)            \(=\dfrac{7}{4}\)

\(x+\dfrac{4}{5}=\dfrac{8}{9}\)

\(x\)          \(=\dfrac{8}{9}-\dfrac{4}{5}\)

\(x\)            \(=\dfrac{4}{45}\)

Mn giúp e với ak

a) \(\sqrt{x^2-6x+9}\)

\(=\sqrt{\left(x^2-2.x.3+3^2\right)}\)

\(=\sqrt{\left(x-3\right)^2}\) ≥0,∀x

⇒x∈\(R\)

b) \(\sqrt{x^2-2x+1}\)

\(=\sqrt{\left(x^2-2.x.1+1^2\right)}\)

\(=\sqrt{\left(x-1\right)^2}\) ≥0,∀x

⇒x∈\(R\)

a, 7x - 3 = 3x + 1

7x - 3x = 1 + 3

4x = 4

x = 4 : 4

x = 1

b, 2x - 5 - 9x = 9

2x - 9x = 9 + 5

-7x = 14

x = 14 : (-7)

x = -2

c, 4 - (x + 3) = 7 - (2x + 6)

4 - x - x = 7 - 2x - 6

-x - x + 2x = 7 - 6 - 4

0x = -3

Vì 0 nhân cho số nào cũng bằng 0 nên không có giá trị nào của x thỏa mãn

7x-3=3x+1

7x-3x=3+1

4x=4

x=4/4

x=1

vậy x =1

b , 2x-5-9x=9

2x-9x-5=9

-7x=9+5

-7x=14

x=14/(-7)

x=-2

vậy x=-2

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

\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)

\(\Leftrightarrow2x=-10\)

hay x=-5

b: Ta có: \(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)

\(\Leftrightarrow6x^2+27x+4x+18-6x^2-x-12x-2=x+1-x+6\)

\(\Leftrightarrow18x+16=7\)

hay \(x=-\dfrac{1}{2}\)

c: Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)

\(\Leftrightarrow3\left(6x^2-2x-3x+1\right)-\left(18x^2-2x-27x+3\right)=0\)

\(\Leftrightarrow18x^2-15x+3-18x^2+27x-3=0\)

hay x=0

a) (x-2)³+6(x+1)²-x³+12=0

\(\Rightarrow\)x³-6x²+12x-8+6(x²+2x+1)-x³+12=0

\(\Rightarrow\)x³-6x²+12x-8+6x²+12x+6-x³+12=0

\(\Rightarrow\)24x+10=0

\(\Rightarrow\)24x=-10

\(\Rightarrow\)x=\(\dfrac{-10}{24}=\dfrac{-5}{12}\)

b)(x-5)(x+5)-(x+3)²+3(x-2)²=(x+1)²-(x-4)(x+4)+3x²

\(\Rightarrow\)x²-25-(x²+6x+9)+3(x²-4x+4)=x²+2x+1-(x²-16)+3x²

\(\Rightarrow\)x²​-25-x²-6x-9+3x²-12x+12=x²+2x+1-x²+16+3x²

\(\Rightarrow\)3x²-18x-22=3x²+2x+17

\(\Rightarrow\)3x²-18x-22-3x²-2x-17=0

\(\Rightarrow\)-20x-39=0

\(\Rightarrow\)-20x=39

\(\Rightarrow\)x=\(-\dfrac{39}{20}\)

c) (2x+3)² +(x-1)(x+1)=5(x+2)²-(x-5)(x+1)+(x+4)

⇒4x²+12x+9+x²-1=5(x²+4x+4)-(x²+x-5x-5)+x+4

⇒5x²+12x+8=5x²+20x+20-x²-x+5x+5+x+4

⇒5x²+12x+8-5x²-20x-20+x²+x-5x-5-x-4=0

⇒x²-13x-21=0

a) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)

\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)

\(\Leftrightarrow24x+25=15\)

\(\Leftrightarrow24x=-10\)

hay \(x=-\dfrac{5}{12}\)

b) Ta có: \(2x^3-50x=0\)

\(\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)

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

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

\(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)

\(\Leftrightarrow x^2+8x-9=0\)

\(\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=1\end{matrix}\right.\)

d) Ta có: \(x^3-x=0\)

\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

e) Ta có: \(27x^3-27x^2+9x-1=1\)

\(\Leftrightarrow\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2-1^3=1\)

\(\Leftrightarrow\left(3x-1\right)^3=1\)

\(\Leftrightarrow3x-1=1\)

\(\Leftrightarrow3x=2\)

hay \(x=\dfrac{2}{3}\)

a)-6x=6

x=-1

b)8x=35->x=35/8

c)8|x|=35->|x|=35/2->x=35/2;x=-35/2

mấy ý kia tương tự bạn ạ!

Các bạn giải bài 2 trước đi,  năn nỉ mà!!!