5(x-2)-125=15

=>\(5\left(x-2\right)=15+125=140\)

=>x-2=28

=>x=28+2=30

Câu trả lời:

5. (𝑥 − 2) − 125 = 15 

5.  ( x - 2)          = 125 - 15

5.  ( x - 2)          = 110

      x - 2            = 110 : 5

       x - 2           =  22

       x                 =  22 - 2

       x                 =  20

2 tiếng rồi chưa bạn nào làm à :v để "Top 4 Battle City" :))

( x + 1 )²( 3x + 2 )( 3x + 4 ) - 8 = 0

<=> ( x² + 2x + 1 )( 9x² + 18x + 8 ) - 8 = 0

Đặt x² + 2x + 1 = y

pt <=> y( 9y - 1 ) - 8 = 0

<=> 9y² - y - 8 = 0

<=> ( y - 1 )( 9y + 8 ) = 0

<=> ( x² + 2x + 1 - 1 )[ 9( x² + 2x + 1 ) + 8 ] = 0

<=> x( x + 2 )[ 9( x + 1 )² + 8 ] = 0

Vì 9( x + 1 )² + 8 ≥ 8 > 0 ∀ x

=> x( x + 2 ) = 0

<=> x = 0 hoặc x = -2

Vậy tập nghiệm của phương trình là S = { 0 ; -2 }

Thanks bạn nhiều nhá!

\(3x\left(x-y\right)+x-y\)

\(=3x\left(x-y\right)+1\left(x-y\right)\)

\(=\left(x-y\right)\left(3x+1\right)\)

\(=\left(x-y\right)\left(3x+1\right)\)

\(1,2\sqrt{27}+5\sqrt{12}-3\sqrt{48}\\ =2.3\sqrt{3}+5.2\sqrt{3}-3.4\sqrt{3}\\ =6\sqrt{3}+10\sqrt{3}-12\sqrt{3}\\ =4\sqrt{3}\)

\(2,\sqrt{147}+\sqrt{75}-4\sqrt{27}\\ =7\sqrt{3}+5\sqrt{3}-4.3\sqrt{3}\\ =7\sqrt{3}+5\sqrt{3}-12\sqrt{3}\\ =\sqrt{3}\left(7+5-12\right)\\ =0\)

\(3,3\sqrt{2}\left(4-\sqrt{2}\right)+3\left(1-2\sqrt{2}\right)^2\\ =3\sqrt{2}.\left(4-\sqrt{2}\right)+3\left(1-4\sqrt{2}+8\right)\\ =12\sqrt{2}-6+3-12\sqrt{2}+24\\ =21\)

\(4,2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}\\ =2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}\\ =\sqrt{5}\left(2-5-4+11\right)\\ =4\sqrt{5}\)

1: =6căn 3+10căn 3-12căn 3=4căn 3

2: =7căn 3+5căn 3-12căn 3=0

3: =12căn 2-6+3(9-4căn 2)

=12căn 2-6+27-12căn 2=21

4: =2căn 5-5căn 5+4căn 5+9 căn 5

=10căn 5

Câu 1: 

a) Ta có: 7x+21=0

\(\Leftrightarrow7x=-21\)

hay x=-3

Vậy: S={-3}

b) Ta có: 3x-2=2x-3

\(\Leftrightarrow3x-2-2x+3=0\)

\(\Leftrightarrow x+1=0\)

hay x=-1

Vậy: S={-1}

c) Ta có: 5x-2x-24=0

\(\Leftrightarrow3x=24\)

hay x=8

Vậy: S={8}

Câu 2: 

a) Ta có: \(\left(2x+1\right)\left(x-1\right)=0\)

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

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

b) Ta có: \(\left(2x-3\right)\left(-x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\-x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\-x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=7\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{3}{2};7\right\}\)

c) Ta có: \(\left(x+3\right)^3-9\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left[\left(x+3\right)^2-9\right]=0\)

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

\(\Leftrightarrow x\left(x+3\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=-6\end{matrix}\right.\)

Vậy: S={0;-3;-6}

Ta có: \(x^5-2x^4+3x^3-4x^2+2\)

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

\(=x^4\left(x-1\right)-x^3\left(x-1\right)+2x^2\left(x-1\right)-2\left(x^2-1\right)\)

\(=\left(x-1\right)\left(x^4-x^3+2x^2\right)-2\left(x-1\right)\left(x+1\right)\)

\(=\left(x-1\right)\left(x^4-x^3+2x^2-2x-2\right)\)

a)\(3x-\dfrac{2}{5}=0=>3x=\dfrac{2}{5}=>x=\dfrac{2}{15}\)

b)\(\left(x-3\right)\left(2x+8\right)=0=>\left[{}\begin{matrix}x-3=0\\2x=-8\end{matrix}\right.=>\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\)

c)\(3x^2-x-4=0=>3x^2+3x-4x-4=0=>\left(3x-4\right)\left(x+1\right)=0\)

\(=>\left[{}\begin{matrix}3x=4\\x+1=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-1\end{matrix}\right.\)

b: Ta có: \(5\left(x-1\right)^2-\left(1-x\right)\)

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

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

\(=\left(x-1\right)\left(5x-4\right)\)

a: Ta có: \(5x^2-4xy-x^2y\)

\(=x\left(5x-4y-xy\right)\)

2:

a: A(x)=0

=>5x-10-2x-6=0

=>3x-16=0

=>x=16/3

b: B(x)=0

=>5x^2-125=0

=>x^2-25=0

=>x=5 hoặc x=-5

c: C(x)=0

=>2x^2-x-3=0

=>2x^2-3x+2x-3=0

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

=>x=3/2 hoặc x=-1

(y + 6x)/y

= (3x + 6x)/(3x)

= (9x)/(3x)

= 3   (1)

y/x = 3x/x = 3   (2)

Từ (1) và (2) suy ra

(y + 6x)/y = y/x (cùng bằng 3)