a: \(\Leftrightarrow12x^2-10x-12x^2-28x=7\)

=>-38x=7

hay x=-7/38

b: \(\Leftrightarrow-10x^2-5x+9x^2+6x+x^2-\dfrac{1}{2}x=0\)

=>1/2x=0

hay x=0

c: \(\Leftrightarrow18x^2-15x-18x^2-14x=15\)

=>-29x=15

hay x=-15/29

d: \(\Leftrightarrow x^2+2x-x-3=5\)

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

\(\text{Δ}=1^2-4\cdot1\cdot\left(-8\right)=33>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-1-\sqrt{33}}{2}\\x_2=\dfrac{-1+\sqrt{33}}{2}\end{matrix}\right.\)

e: \(\Leftrightarrow-15x^2+10x-10x^2-5x-5x=4\)

\(\Leftrightarrow-25x^2=4\)

\(\Leftrightarrow x^2=-\dfrac{4}{25}\left(loại\right)\)

a/ \(x^2\left(x-5\right)+5-x=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\x-5=0\end{matrix}\right.\)

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

Vậy...

b/ \(3x^4-9x^3=-9x^2+27x\)

\(\Leftrightarrow3x^4-9x^3+9x^2-27x=0\)

\(\Leftrightarrow3x^3\left(x-3\right)+9x\left(x-3\right)=0\)

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

\(\Leftrightarrow3x\left(x-3\right)\left(x^2+3\right)=0\)

Vì \(x^2+3>0\forall x\)

\(\Leftrightarrow3x\left(x-3\right)=0\)

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

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

Vậy..

c/ \(x^2\left(x+8\right)+x^2=-8x\)

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

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

\(\Leftrightarrow x\left(x+8\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+8=0\\x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\\x=-1\end{matrix}\right.\)

Vậy...

d/ \(\left(x+3\right)\left(x^2-3x+5\right)=x^2+3x\)

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

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

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

Vì \(\left(x-2\right)^2+1>0\forall x\)

\(\Leftrightarrow x+3=0\Leftrightarrow x=-3\)

Vậy..

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

\(b,3x^4-9x^3=-9x^2+27x\\ \Leftrightarrow3x^4-9x^3+9x^2-27x=0\\ \Leftrightarrow3x^3\left(x-3\right)+9x\left(x-3\right)=0\\ \Leftrightarrow3x\left(x-3\right)\left(x^2+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

do \(x^2+2>0\)

\(c,x^2\left(x+8\right)+x^2=-8x\\ \Leftrightarrow x^2\left(x+8\right)+x^2+8x=0\\ \Leftrightarrow x^2\left(x+8\right)+x\left(x+8\right)=0\\ \Leftrightarrow x\left(x+1\right)\left(x+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-8\end{matrix}\right.\)

\(d,\left(x+3\right)\left(x^2-3x+5\right)=x^2+3x\\ \Leftrightarrow\left(x+3\right)\left(x^2-3x+5\right)-x^2-3x=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-3x+5\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-3x+5-x\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x^2-4x+5=0\end{matrix}\right.\)

\(\Delta=b^2-4ac=\left(-4\right)^2-4\cdot1\cdot5=16-20=-4< 0\)\(\rightarrow\)\(x^2-4x+5\) vô nghiệm

Vậy \(x=-3\)

\(a.\frac{4x-8}{2x^2+1}=0\\ \Leftrightarrow4x-8=0\\ \Leftrightarrow4\left(x-2\right)=0\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\)

Vậy nghiệm của phương trình trên là \(2\)

\(b.\frac{x^2-x-6}{x-3}=0\left(x\ne3\right)\\\Leftrightarrow x^2-x-6=0\\ \Leftrightarrow x^2+2x-3x-6=0\\\Leftrightarrow x\left(x+2\right)-3\left(x+2\right)=0\\\Leftrightarrow \left(x-3\right)\left(x+2\right)=0\\\Leftrightarrow \left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\left(ktm\right)\\x=-2\left(tm\right)\end{matrix}\right.\)

Vậy nghiệm của phương trình trên là \(-2\)

\(c.\frac{x+5}{3x-6}-\frac{1}{2}=\frac{2x-3}{2x-4}\left(x\ne2\right)\\ \Leftrightarrow\frac{x+5}{3\left(x-2\right)}-\frac{1}{2}=\frac{2x-3}{2\left(x-2\right)}\\\Leftrightarrow \frac{2\left(x+5\right)}{6\left(x-2\right)}-\frac{3\left(x-2\right)}{6\left(x-2\right)}=\frac{3\left(2x-3\right)}{6\left(x-2\right)}\\\Leftrightarrow 2\left(x+5\right)-3\left(x-2\right)=3\left(2x-3\right)\\\Leftrightarrow 2x+10-3x+6=6x-9\\\Leftrightarrow 2x-3x-6x=-10-6-9\\\Leftrightarrow -7x=-25\\\Leftrightarrow x=\frac{25}{7}\left(tm\right)\)

Vậy nghiệm của phương trình trên là \(\frac{25}{7}\)

\(1,\dfrac{x-2}{2}=3.\dfrac{1-3x}{6}\\ \Leftrightarrow\dfrac{x-2}{2}=\dfrac{1-3x}{2}\\ \Leftrightarrow x-2=1-3x\\ \Leftrightarrow4x=3\\ \Leftrightarrow x=\dfrac{3}{4}\)

2, mik có sửa đề vì đề của bn sai

ĐKXĐ:\(x\ne\pm\dfrac{1}{3}\)

\(\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}=\dfrac{5}{1-9x^2}\\ \Leftrightarrow\dfrac{\left(1-3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}-\dfrac{\left(1+3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}-\dfrac{5}{\left(1-3x\right)\left(1+3x\right)}=0\\ \Leftrightarrow\dfrac{1-6x+9x^2-1-6x-9x^2-5}{\left(1+3x\right)\left(1-3x\right)}=0\\ \Rightarrow-12x-5=0\\ \Leftrightarrow x=-\dfrac{5}{12}\left(tm\right)\)

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

=>3(3x+2)-4(3x+1)=10

=>9x+6-12x-4=10

=>-3x+2=10

=>-3x=8

=>x=-8/3

b: \(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{9x-10}{4-x^2}\)

=>(x-1)(x-2)-x(x+2)=-9x+10

=>x^2-3x+2-x^2-2x=-9x+10

=>-5x+2=-9x+10

=>x=2(loại)

a) \(\left(x+2\right)\left(x^2-24+4\right)\left(x^3+8\right)\)

\(=\left(x+2\right)\left(x^2-20\right)\left(x^3+8\right)\)

\(=\left(x^3-20x+2x^2-40\right)\left(x^3+8\right)\)

\(=x^6+8x^3-20x^4+160x+2x^5+16x^2-40x^3-120\)

\(=x^6+2x^5-20x^4-32x^3+16x^2+160x-120\)

b) \(\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)

\(=8x^3+2x^2+\dfrac{1}{2}x-2x^2-\dfrac{1}{2}x-\dfrac{1}{8}\)

\(=8x^3-\dfrac{1}{8}\)

c) \(\left(x^2+y\right)\left(x^2-y\right)+y^2+x^4\)

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

\(=x^4-y^2+y^2+x^4\)

\(=2x^4\)

d) \(\left(x+3\right)\left(x^2-3x+9\right)-x^3\)

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

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

\(=27\)

e) \(\left(3x+y\right)\left(9x^2-3xy+y^2\right)-26x^3\)

\(=\left(3x+y\right)\left[\left(3x\right)^2-3x\cdot y+y^2\right]-26x^3\)

\(=\left(3x\right)^3+y^3-26x^3\)

\(=27x^3+y^3-26x^3\)

\(=x^3+y^3\)

g) \(\left(x+3y\right)\left(x^2-3xy+9y^2\right)+\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)

\(=\left(x+3y\right)\left[x^2-x\cdot3y+\left(3y\right)^2\right]+\left(3x-y\right)\left[\left(3x\right)^2+3x\cdot y+y^2\right]\)

\(=\left[x^3+\left(3y\right)^3\right]+\left[\left(3x\right)^3-y^3\right]\)

\(=x^3+27y^3+27x^3-y^3\)

\(=28x^3+26y^3\)

Yêu cầu của đề là gì ?

a) Sửa đề:

(x + 2)(x² - 2x + 4)(x³ + 8)

= (x³ + 8)(x³ + 8)

= (x³ + 8)²

b) (2x - 1/2)(4x² + x + 1/4)

= (2x)³ - (1/2)³

= 8x³ - 1/8

c) (x² + y)(x² - y) + y² + x⁴

= (x²)² - y² + y² + x⁴

= 2x⁴

d) (x + 3)(x² - 3x + 9) - x³

= x³ + 3³ - x³

= 27

e) (3x + y)(9x² - 3xy + y²) - 26x³

= (3x)³ + y³ - 26x³

= 27x³ + y³ - 26x³

= x³ + y³

g) (x + 3y)(x² - 3xy + 9y²) + (3x - y)(9x² + 3xy + y²)

= x³ + (3y)³ + (3x)³ - y³

= x³ + 27y³ + 27x³ - y³

= 28x³ + 26y³

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