Bài 1.

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

\(\Leftrightarrow x^3-1-x^3-x=4\)

\(\Leftrightarrow-x=5\)

hay x=-5

c: Ta có: \(\left(2x-1\right)^3+\left(x+2\right)^3-9x\left(x+1\right)\left(x-1\right)=7\)

\(\Leftrightarrow8x^3-12x^2+6x-1+x^3+6x^2+12x+8-9x^3+9x=7\)

\(\Leftrightarrow-6x^2+27x=0\)

\(\Leftrightarrow-3x\left(2x-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{9}{2}\end{matrix}\right.\)

a) \(\Rightarrow9x^2+24x+16-9x^2+1=49\)

\(\Rightarrow24x=32\Rightarrow x=\dfrac{4}{3}\)

b) \(\Rightarrow x^2-13x+22=0\)

\(\Rightarrow\left(x-11\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=11\\x=2\end{matrix}\right.\)

c) \(\Rightarrow x^2-3x-10=0\)

\(\Rightarrow\left(x-5\right)\left(x+2\right)=0\)

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

a) \(\left(x-1\right)^3\)

\(=x^3-3x^2+3x-1\)

b) \(\left(2x-3y\right)^3\)

\(=\left(2x\right)^3-3\left(2x\right)^23y+3.2x\left(3y\right)^3+\left(3y\right)^3\)

\(=8x^3-36x^2y+54xy^2-27y^3\)

Bài 3: 

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

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+6x^2=5\)

\(\Leftrightarrow12x=13\)

hay \(x=\dfrac{13}{12}\)

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

\(\Leftrightarrow x^3-1-x^3+4x=4\)

\(\Leftrightarrow4x=5\)

hay \(x=\dfrac{5}{4}\)

a) x² - 2 = 0

x² = 2

x = -√2 (loại) hoặc x = √2 (loại)

Vậy không tìm được x Q thỏa mãn đề bài

b) x² + 7/4 = 23/4

x² = 23/4 - 7/4

x² = 4

x = 2 (nhận) hoặc x = -2 (nhận)

Vậy x = -2; x = 2

c) (x - 1)² = 0

x - 1 = 0

x = 1 (nhận)

Vậy x = 1

bạn linh này giỏi dữ ta

a) x² - 2 = 0

x² = 2

x = √2 hoặc -√2 (loại) (x ϵ Q)

Vậy x ϵ rỗng

b) x² + 7/4 =23/4

x² = 23/4 - 7/4

x² = 16/4 = 4

x² = 4 = (-2)² = 2²

x² = (-2)²

x = -2 (Nhận)

x² = 2

x = 2 (Nhận)

Vậy x ϵ ( 2 , -2 )

c) (x-1)² = 0

x-1 = 0

x = 1

Vậy x = 1

\(a,\Leftrightarrow6x^2-6x^2-11x+10=-12\\ \Leftrightarrow-11x=-22\\ \Leftrightarrow x=2\\ b,\Leftrightarrow x^3+27-x^3-2x=12-5x\\ \Leftrightarrow3x=-15\\ \Leftrightarrow x=-5\\ c,\Leftrightarrow x^2-6x-16=0\\ \Leftrightarrow\left(x-8\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

a: ta có: \(6x^2-\left(2x+5\right)\left(3x-2\right)=-12\)

\(\Leftrightarrow6x^2-6x^2+4x-15x+10=-12\)

\(\Leftrightarrow-11x=-22\)

hay x=2

b: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2+2\right)=12-5x\)

\(\Leftrightarrow x^3+27-x^3-2x+5x=12\)

\(\Leftrightarrow x=-5\)

a: =>2x^2=4

=>x^2=2

=>\(x=\pm\sqrt{2}\)

b: =>(x+1)^2-4=0

=>(x+1+2)(x+1-2)=0

=>(x+3)(x-1)=0

=>x=1 hoặc x=-3

c: =>(2x-1)^2-3^2=0

=>(2x-1-3)(2x-1+3)=0

=>(2x-4)(2x+2)=0

=>x=2 hoặc x=-1

d: x^2-x=0

=>x(x-1)=0

=>x=0 hoặc x=1

a) Đặt: \(A=1+2^2+2^3+...+2^{10}\)

\(\Rightarrow2A=2\left(1+2^2+2^3+...+2^9+2^{10}\right)\)

\(\Rightarrow2A=2+2^3+2^4+...+2^{10}+2^{11}\)

\(\Rightarrow2A-A=\left(2+2^3+2^4+...+2^{10}+2^{11}\right)-\left(1+2^2+2^3+...+2^{10}\right)\)

\(\Rightarrow A=\left(2^3-2^3\right)+\left(2^4-2^4\right)+...+\left(2-1\right)+\left(2^{11}-2^2\right)\)

\(\Rightarrow A=0+0+...+1+\left(2^{11}-2^2\right)\)

\(\Rightarrow A=1+2^{11}-2^2=1+2048-4=2045\)

Vậy: \(1+2^2+2^3+...+2^{10}=2045\)

b) 

a] \(60-3\left(x-1\right)=2^3\cdot3\)

\(\Rightarrow60-3\left(x-1\right)=24\)

\(\Rightarrow3\left(x-1\right)=36\)

\(\Rightarrow x-1=12\)

\(\Rightarrow x=13\)

b] \(\left(3x-2\right)^3=2\cdot2^5\)

\(\Rightarrow\left(3x-2\right)^3=2^6\)

\(\Rightarrow\left(3x-2\right)^3=\left(2^2\right)^3\)

\(\Rightarrow3x-2=2^2\)

\(\Rightarrow3x=6\)

\(x=2\)

c] \(5^{x+1}-5^x=500\)

\(\Rightarrow5^x\left(5-1\right)=500\)

\(\Rightarrow5^x\cdot4=500\)

\(\Rightarrow5^x=125\)

\(\Rightarrow5^x=5^3\)

\(\Rightarrow x=3\)

d] \(x^2=x^4\)

\(\Rightarrow x=x^2\)

\(\Rightarrow x-x^2=0\)

\(\Rightarrow x\left(1-x\right)=0\)

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

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

giúp mình đi các bạn

tách ra

\(1\)) \(5-\left(10-x\right)=7\)
\(10-x=5-7\)
\(10-x=-2\)
\(x=10-\left(-2\right)\)
\(x=12\)

\(2\)) \(-32-\left(x-5\right)=0\)
\(x-5=-32-0\)
\(x-5=-32\)
\(x=-32+5\)
\(x=-27\)

1)10-x=5-7

x=10-(-2)

x=12

2)x+5=0+32

x=32-5

x=27

3) -12+2x-9+x=0

-12+(2x+x-9)=0

3x-9=0+12

3x=12+9

x=21:3

x=7

4) 15-x=1-11

x=15-(-10)

x=25

5) 4-(27+3)=x-9

4-30=x-9

-26+9=x

x=-17