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

\(\Leftrightarrow\left(x+2\right)\left(x^2-4x+4\right)-x^2\left(x+2\right)=5\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-4x+4-x^2\right)=5\)

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

\(\Leftrightarrow4x-4x^2+8-8x=5\)

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

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

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

\(\Leftrightarrow2x\left(2x-1\right)+3\left(2x-1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x+3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{3}{2}\end{matrix}\right.\)

Vậy \(x=\left\{\frac{1}{2};-\frac{3}{2}\right\}\)

b) \(6x^2-6x\left(-2+x\right)=36\)

\(\Leftrightarrow6x^2+12x-6x^2=36\)

\(\Leftrightarrow12x=36\)

\(\Leftrightarrow x=3\)

Vậy x = 3

c) \(\left(x+2\right)^2+\left(x-3\right)^2-2\left(x-1\right)\left(x+1\right)=9\)

\(\Leftrightarrow x^2+4x+4+x^2-6x+9-2\left(x^2-1\right)=9\)

\(\Leftrightarrow2x^2-2x+13-2x^2+2=9\)

\(\Leftrightarrow15-2x=9\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=3\)

Vậy x = 3

d) \(\left(x+5\right)^2-9=0\)

\(\Leftrightarrow\left(x+5\right)^2=9\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x+5\right)^2=3^2\\\left(x+5\right)^2=\left(-3\right)^2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x+5=3\\x+5=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-8\end{matrix}\right.\)

Vậy x ={-2; -8}

e) \(\left(x-2\right)^3=x^3+6x^2=7\) (Câu này sai đề thì phải! Mình sửa lại đề, có gì không giống với đề của bạn thì ib mình sửa nha!)

\(\left(x-2\right)^3-x^3+6x^2=7\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+6x^2=7\)

\(\Leftrightarrow12x-8=7\)

\(\Leftrightarrow12x=15\)

\(\Leftrightarrow x=\frac{5}{4}\)

Vậy \(x=\frac{5}{4}\)

#Chúc bạn học tốt!

\(\frac{2x-3}{\left(7-6x\right)^2}+\frac{x-2}{\left(7-6x\right)^2}=\frac{6x-3}{\left(3x-5\right)^2}-\frac{12x-10}{\left(3x-5\right)^2}\)

\(\Leftrightarrow\frac{2x-3+x-2}{\left(7-6x\right)^2}=\frac{6x-3-12x+10}{\left(3x-5\right)^2}\)

\(\Leftrightarrow\frac{3x-5}{\left(7-6x\right)^2}=\frac{7-6x}{\left(3x-5\right)^2}\)

\(\Leftrightarrow\left(7-6x\right)^3=\left(3x-5\right)^3\)

\(\Leftrightarrow7-6x=3x-5\)

\(\Leftrightarrow7+5=3x+6x\)

\(\Leftrightarrow12=9x\)

\(\Leftrightarrow x=\frac{4}{3}\)

Vậy \(x=\frac{4}{3}\)

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

\(\Leftrightarrow3-x^2+6-2x=0\)

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

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

\(\Leftrightarrow x^2+2x+1=10\)

\(\Leftrightarrow\left(x+1\right)^2=10\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=\sqrt{10}\\x+1=-\sqrt{10}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{10}-1\\x=-\sqrt{10}-1\end{matrix}\right.\)

Vậy: \(S=\left\{\sqrt{10}-1;-\sqrt{10}-1\right\}\)

2) Ta có: \(5\left(2x-1\right)+7=4\left(2-x\right)+2\)

\(\Leftrightarrow10x-5+7=8-4x+2\)

\(\Leftrightarrow10x+4x=8+2+5-7\)

\(\Leftrightarrow14x=8\)

\(\Leftrightarrow x=\dfrac{4}{7}\)

Vậy: \(S=\left\{\dfrac{4}{7}\right\}\)

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

\(\Leftrightarrow x\left(x-6\right)+4\left(x-6\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

Vậy: S={6;-4}

a) \(\left(3x-5\right)\left(2x-1\right)-\left(x+2\right)\left(6x-1\right)=0\)

⇔ \(6x^2-13x+5-6x^2-11x+2=0\)

⇔ \(24x=7\)⇔\(x=\frac{7}{24}\)

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

⇔ \(9x^2-4-9x^2+6x-1=5\)

⇔ \(6x=10\)⇔ \(x=\frac{5}{3}\)

c) \(x^2=-6x-8\)⇔\(x^2+6x+8=0\)⇔\(\left(x+2\right)\left(x+4\right)=0\)

⇔\(\left[{}\begin{matrix}x=-2\\x=-4\end{matrix}\right.\)

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`

Rảnh rỗi thật sự .-.

2x³( 2x - 3 ) - x²( 4x² - 6x + 2 ) = 0

<=> 4x⁴ - 6x³ - 4x⁴ + 6x³ - 2x² = 0

<=> -2x² = 0

<=> x = 0

6x² - ( 2x + 5 )( 3x - 2 ) = 7

<=> 6x² - ( 6x² + 11x - 10 ) = 7

<=> 6x² - 6x² - 11x + 10 = 7

<=> -11x + 10 = 7

<=> -11x = -3

<=> x = 3/11

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

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

\(-2x^2=0\)

\(x^2=0\)

\(x=0\)

\(6x^2-\left(2x+5\right)\left(3x-2\right)=7\)

\(6x^2-\left(6x^2-4x+15x-10\right)=7\)

\(6x^2-\left(6x^2+11x-10\right)=7\)

\(6x^2-6x^2-11x+10=7\)

\(-11x+10=7\)

\(-11x=7-10\)

\(-11x=-3\)

\(x=\left(-3\right)\div\left(-11\right)\)

\(x=\frac{3}{11}\)

a) \(2x^3.\left(2x-3\right)-x^2.\left(4x^2-6x+2\right)=0\)

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

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

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

Vậy \(x=0\)

b) \(6x^2-\left(2x+5\right)\left(3x-2\right)=7\)

\(\Leftrightarrow6x^2-\left(6x^2+11x-10\right)=7\)

\(\Leftrightarrow6x^2-6x^2-11x+10=7\)

\(\Leftrightarrow-11x+10=7\)

\(\Leftrightarrow11x=10-7\)

\(\Leftrightarrow11x=3\)\(\Leftrightarrow x=\frac{3}{11}\)

Vậy \(x=\frac{3}{11}\)

5.

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\frac{5}{6}\left[\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\right]\)

\(\Leftrightarrow1-3sin^2x.cos^2x=\frac{5}{6}\left(1-2sin^2x.cos^2x\right)\)

\(\Leftrightarrow1-\frac{3}{4}sin^22x=\frac{5}{6}\left(1-\frac{1}{2}sin^22x\right)\)

\(\Leftrightarrow\frac{1}{3}sin^22x=\frac{1}{6}\)

\(\Leftrightarrow sin^22x=\frac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=\frac{\sqrt{2}}{2}\\sin2x=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+k\pi\\x=\frac{3\pi}{8}+k\pi\\x=-\frac{\pi}{8}+k\pi\\x=\frac{5\pi}{8}+k\pi\end{matrix}\right.\)

6.

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+\frac{1}{2}sinx.cosx=0\)

\(\Leftrightarrow1-3sin^2x.cos^2x+\frac{1}{2}sinx.cosx=0\)

\(\Leftrightarrow1-\frac{3}{4}sin^22x+\frac{1}{4}sin2x=0\)

\(\Leftrightarrow-3sin^22x+sin2x+4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=-1\\sin2x=\frac{4}{3}>1\left(l\right)\end{matrix}\right.\)

\(\Rightarrow2x=-\frac{\pi}{2}+k2\pi\)

\(\Rightarrow x=-\frac{\pi}{4}+k\pi\)

1.

\(\Rightarrow4cos^3x.cos3x+4sin^3x.sin3x=\sqrt{2}\)

\(\Leftrightarrow\left(3cosx+cos3x\right)cos3x+\left(3sinx-sin3x\right)sin3x=\sqrt{2}\)

\(\Leftrightarrow3\left(cos3x.cosx+sin3x.sinx\right)+cos^23x-sin^23x=\sqrt{2}\)

\(\Leftrightarrow3cos2x+cos6x=\sqrt{2}\)

\(\Leftrightarrow3cos2x+4cos^32x-3cos2x=\sqrt{2}\)

\(\Leftrightarrow4cos^32x=\sqrt{2}\)

\(\Leftrightarrow cos2x=\frac{\sqrt{2}}{2}\)

\(\Rightarrow\left[{}\begin{matrix}2x=\frac{\pi}{4}+k2\pi\\2x=-\frac{\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+k\pi\\x=-\frac{\pi}{8}+k\pi\end{matrix}\right.\)

giải đc sao pn dễ mk

chẳng ai giải, thôi mình giải vậy!

a) Đặt \(y=x^2+4x+8\),phương trình có dạng:

\(t^2+3x\cdot t+2x^2=0\)

\(\Leftrightarrow t^2+xt+2xt+2x^2=0\)

\(\Leftrightarrow t\left(t+x\right)+2x\left(t+x\right)=0\)

\(\Leftrightarrow\left(2x+t\right)\left(t+x\right)=0\)

\(\Leftrightarrow\left(2x+x^2+4x+8\right)\left(x^2+4x+8+x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-4\end{cases}}\)vậy tập nghiệm của phương trình là:S={-2;-4}

b) nhân 2 vế của phương trình với 12 ta được:

\(\left(6x+7\right)^2\left(6x+8\right)\left(6x+6\right)=72\)

Đặt y=6x+7, ta được:\(y^2\left(y+1\right)\left(y-1\right)=72\)

giải tiếp ra ta sẽ được S={-2/3;-5/3}

c) \(\left(x-2\right)^4+\left(x-6\right)^4=82\)

S={3;5}

d)s={1}

e) S={1;-2;-1/2}

f) phương trình vô nghiệm