a

4x²--25=0

=> (2x)²2 --5²  =0

=> (2x-5)(2x+5)=0

\(\orbr{\begin{cases}2x-5=0\\2x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}X=\frac{5}{2}\\X=\frac{-5\:\:. \:\:\:\:\:\:\:\:\:\:TT}{2}\end{cases}Mình\:}\)

\(4x^2=25\Rightarrow x^2=\frac{25}{4}\Rightarrow x=\sqrt{\frac{25}{4}}\) \(=\frac{5}{2}\)

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

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

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

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

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

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

=\(\left(x-1\right)\left(x^4-4\right)\) = 0

=> \(x-1=0\) hoặc  \(x^4-4=0\)

=> \(x=1\) hoặc \(x=\pm\sqrt{2}\)

câu 2

a)\(\left(3x^2\right)^3-\left(2x\right)^3\)

= \(\left(3x^2-2x\right)\left(9x^4-54x^5+36x^4-4x^2\right)\)

= \(x\left(3x-2\right)\left(9x^4-54x^5+36x^4-4x^2\right)\)

may be wrong , but chawsc k nhiều , chỗ nào k hiểu ib hỏi mk sai nha  <3

a , ( 2x - 5 ) ( 2x + 5 ) = 0 .... tự làm nhé
 

1, 

a, \(\left(2x-5\right)\cdot\left(2x+5\right)=0\)

\(x=\frac{5}{2}\)

x\(=-\frac{5}{2}\)

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

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

x=2

x=2/3

2, 

a (3x^2)^3-(2x)^3

(3x^2-2x)(9x^4+6x^3+4x^2)

\(4x^2-25=0\)

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

\(\Rightarrow\orbr{\begin{cases}2x-5=0\\2x+5=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{5}{2}\end{cases}}\)

Vậy \(\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{5}{2}\end{cases}}\)

\(27x^6-8x^3=\left(3x^2\right)^3-\left(2x\right)^3=\left(3x^2-2x\right)\left[\left(3x^2\right)^2+3x^2.2x+\left(2x\right)^2\right]=x^3.\left(3x-2\right).\left(3x^2+6x+4\right)\)

a) 4x² - 25 = 0

    4x²            = 25

    ( 2x)²     =  5²

     2x        = 5

=>  x         = 5/2

1a) 4x² - 25 = 0 =>  4x² = 25 => x² = \(\frac{25}{4}\)= \(\left(\frac{5}{2}\right)^2\)=> x = \(\frac{5}{2}\)

1) Ta có: \(\left(x^2-4x+4\right)\left(x^2+4x+4\right)-\left(7x+4\right)^2=0\)

\(\Leftrightarrow\left(x-2\right)^2\cdot\left(x+2\right)^2-\left(7x+4\right)^2=0\)

\(\Leftrightarrow\left[\left(x-2\right)\left(x+2\right)\right]^2-\left(7x+4\right)^2=0\)

\(\Leftrightarrow\left(x^2-4\right)^2-\left(7x+4\right)^2=0\)

\(\Leftrightarrow\left(x^2-4-7x-4\right)\left(x^2-4+7x+4\right)=0\)

\(\Leftrightarrow\left(x^2-7x-8\right)\left(x^2+7x\right)=0\)

\(\Leftrightarrow x\left(x+7\right)\left(x^2-8x+x-8\right)=0\)

\(\Leftrightarrow x\left(x+7\right)\left[x\left(x-8\right)+\left(x-8\right)\right]=0\)

\(\Leftrightarrow x\left(x+7\right)\left(x-8\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+7=0\\x-8=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-7\\x=8\\x=-1\end{matrix}\right.\)

Vậy: S={0;-7;8;-1}

2) Ta có: \(x^3-8x^2+17x-10=0\)

\(\Leftrightarrow x^3-2x^2-6x^2+12x+5x-10=0\)

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

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

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

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

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

Vậy: S={2;1;5}

3) Ta có: \(2x^3-5x^2-x+6=0\)

\(\Leftrightarrow2x^3-4x^2-x^2+2x-3x+6=0\)

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

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

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

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

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

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

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

4) Ta có: \(4x^4-4x^2-3=0\)

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

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

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

mà \(2x^2+1>0\forall x\in R\)

nên \(2x^2-3=0\)

\(\Leftrightarrow2x^2=3\)

\(\Leftrightarrow x^2=\frac{3}{2}\)

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

Vậy: \(S=\left\{\sqrt{\frac{3}{2}};-\sqrt{\frac{3}{2}}\right\}\)

#)Giải :

Bài 1 :

a) \(9\left(4x+3\right)^2=16\left(3x-5\right)^2\)

\(\Leftrightarrow144x^2+216x+81=144x^2-480x+400\)

\(\Leftrightarrow144x^2+216=144x^2-480x+319\)

\(\Leftrightarrow696x=319\)

\(\Leftrightarrow x=\frac{11}{24}\)

b) \(\left(x^3-x^2\right)^2-4x^2+8x-4=0\)

\(\Leftrightarrow\left(x-1\right)^2\left(x^2+2\right)\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)=0\)

\(\Leftrightarrow x=1\)

c) \(x^5+x^4+x^3+x^2+x+1=0\)

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

\(\Leftrightarrow x=-1\)

a) 9(4x + 3)² = 16(3x - 5)²

=> [3(4x + 3)]² - [4(3x - 5)]² = 0

=> (12x + 9)² - (12x - 20)² = 0

=> (12x + 9 - 12x + 20)(12x + 9 + 12x - 20) = 0

=> 29.(24x - 11) = 0

=> 2x - 11 = 0

=> 2x = 11

=>  x = 11 : 2 = 11/2

b) (x³ - x²)² - 4x² + 8x - 4 = 0

=> (x³ - x²)² - (2x - 2)² = 0

=> (x³ - x² - 2x + 2)(x³ - x² + 2x - 2) = 0

=> [x²(x - 1) - 2(x - 1)][x²(x - 1) + 2(x - 1)] = 0

=> (x² - 2)(x - 1)(x² + 2)(x - 1) = 0

=> (x² - 2)(x² + 2)(x - 1)² = 0

=> x² - 2 = 0

hoặc : x² + 2 = 0

hoặc : (x - 1)² = 0

=> x² = 2

 hoặc : x² = -2 (vl)

hoặc : x - 1 = 0

=> \(\orbr{\begin{cases}x=\sqrt{2}\\x=-\sqrt{2}\end{cases}}\)

hoặc : x = 1

Vậy ...

c) x⁵  + x⁴ + x³ + x² + x + 1 = 0

=> x⁴(x +1) + x²(x + 1) + (x + 1) = 0

=> (x⁴ + x² + 1)(x + 1) = 0

=> \(\orbr{\begin{cases}x^4+x^2+1=0\\x+1=0\end{cases}}\)

=> \(\orbr{\begin{cases}x^4+x^2=-1\left(vl\right)\\x=-1\end{cases}}\) (vì x⁴ \(\ge\)0 \(\forall\)x; x² \(\ge\)0 \(\forall\)x => x⁴ + x² \(\ge\)0 \(\forall\)x)

=> x = -1

a) 

\(-4x\left(-2x+1\right):-4x-\left(x+2\right)=8\) 

\(-2x+1-x-2=8\) 

\(-3x-1=8\) 

\(-3x=9\) 

\(x=-3\) 

b) 

\(-\frac{1}{2}x^2\left(-4x^2+6x-2\right):\left(\frac{-1}{2}x^2\right)+4\left(x^2-2x+1\right)==0\) 

\(-4x^2+6x-2+4x^2-8x+4=0\) 

\(-2x+2=0\) 

\(-2x=-2\) 

\(x=1\)

ggaao ko

8x²+30x+7=0

 8x²+16x+14x+7=0

8x(x+2) +7(x+2)=0

(8x+7)(x+2)=0

=>\(\orbr{\begin{cases}8x+7=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{7}{8}\\x=-2\end{cases}}}\)

a)

4x²-8x+4=2(1-x)(x+1)

4x²-8x+4-2+2x²=0

6x²-8x+2=0

2(3x²-4x+1)=0

3x²-3x-x+1=0

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

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

\(\Rightarrow\orbr{\begin{cases}3x-1=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=1\end{cases}}}\)

a) ( 5 - 2x )( 2x + 7 ) - 4x² + 25 = 0

<=> ( 5 - 2x )( 2x + 7 ) + ( 5 - 2x )( 5 + 2x ) = 0

<=> ( 5 - 2x )( 2x + 7 + 5 + 2x ) = 0

<=> ( 5 - 2x )( 4x + 12 ) = 0

<=> \(\orbr{\begin{cases}5-2x=0\\4x+12=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-3\end{cases}}\)

b) ( 5x² + 3x - 2 )² - ( 4x² - x - 5 )² = 0 ( như này chứ nhỉ ? )

<=> [ ( 5x² + 3x - 2 ) - ( 4x² - x - 5 ) ][ ( 5x² + 3x - 2 ) + ( 4x² - x - 5 ) ] = 0

<=> ( 5x² + 3x - 2 - 4x² + x + 5 )( 5x² + 3x - 2 + 4x² - x - 5 ) = 0

<=> ( x² + 4x + 3 )( 9x² + 2x - 7 ) = 0

<=> ( x² + x + 3x + 3 )( 9x² + 9x - 7x - 7 ) = 0

<=> [ x( x + 1 ) + 3( x + 1 ) ][ 9x( x + 1 ) - 7( x + 1 ) ] = 0

<=> ( x + 1 )( x + 3 )( x + 1 )( 9x - 7 ) = 0

<=> ( x + 1 )²( x + 3 )( 9x - 7 ) = 0

<=> x + 1 = 0 hoặc x + 3 = 0 hoặc 9x - 7 = 0

<=> x = -1 hoặc x = -3 hoặc x = 7/9

c) 15x⁴ - 8x³ - 14x² - 8x + 15 = 0

<=> 15x⁴ + 22x³ - 30x³ + 15x² + 15x² - 44x² - 30x + 22x + 15 = 0

<=> ( 15x⁴ + 22x³ + 15x² ) - ( 30x³ + 44x² + 30x ) + ( 15x² + 22x + 15 ) = 0

<=> x²( 15x² + 22x + 15 ) - 2x( 15x² + 22x + 15 ) + ( 15x² + 22x + 15 ) = 0

<=> ( 15x² + 22x + 15 )( x² - 2x + 1 ) = 0

<=> ( 15x² + 22x + 15 )( x - 1 )² = 0

<=> \(\orbr{\begin{cases}15x^2+22x+15=0\\\left(x-1\right)^2=0\end{cases}}\)

+) ( x - 1 )² = 0 <=> x = 1

+) 15x² + 22x + 15 = 15( x² + 22/15x + 121/225 ) + 104/15 = 15( x + 11/25 )² + 104/15 ≥ 104/15 > 0 ∀ x

Vậy phương trình có nghiệm duy nhất là x = 1

Cảm ơn bạn câu b thiếu cái mũ 2 sorry :))