a) x² + 10x + 16 = 0

<=> x² + 2x + 8x + 16 = 0

<=> x( x + 2 ) + 8( x + 2 ) = 0

<=> ( x + 2 )( x + 8 ) = 0

<=> \(\orbr{\begin{cases}x+2=0\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-8\end{cases}}\)

b) 4x² - 12x - 7 = 0

<=> 4x² + 2x - 14x - 7 = 0

<=> 2x( 2x + 1 ) - 7( 2x + 1 ) = 0

<=> ( 2x + 1 )( 2x - 7 ) = 0

<=> \(\orbr{\begin{cases}2x+1=0\\2x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{7}{2}\end{cases}}\)

a. \(x^2+10x+16=0\)

\(\Leftrightarrow x^2+8x+2x+16=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-8\end{cases}}\)

b. \(4x^2-12x-7=0\)

\(\Leftrightarrow4x^2+2x-14x-7=0\)

\(\Leftrightarrow2x\left(2x+1\right)-7\left(2x+1\right)=0\)

\(\Leftrightarrow\left(2x-7\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-7=0\\2x+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}2x=7\\2x=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}\)

             Bài làm :

 \(\text{a) }x^2+10x+16=0\)

\(\Leftrightarrow x^2+8x+2x+16=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-8\end{cases}}\)

 \(\text{b) }4x^2-12x-7=0\)

\(\Leftrightarrow4x^2+2x-14x-7=0\)

\(\Leftrightarrow2x\left(2x+1\right)-7\left(2x+1\right)=0\)

\(\Leftrightarrow\left(2x-7\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-7=0\\2x+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}2x=7\\2x=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}.}\)

a) \(\dfrac{3x-4}{2x+5}=\dfrac{3x+7}{2x-20}\left(đk:x\ne-\dfrac{5}{2},x\ne10\right)\)

\(\Rightarrow\left(3x-4\right)\left(2x-20\right)=\left(3x+7\right)\left(2x+5\right)\)

\(\Rightarrow6x^2-68x+80=6x^2+29x+35\)

\(\Rightarrow97x=45\Rightarrow x=\dfrac{45}{97}\)

b) \(\dfrac{10x-5}{7x+2}=\dfrac{50x+10}{35x-29}\left(đk:x\ne-\dfrac{2}{7},x\ne\dfrac{29}{35}\right)\)

\(\Rightarrow\left(10x-5\right)\left(35x-29\right)=\left(50x+10\right)\left(7x+2\right)\)

\(\Rightarrow350x^2-465x+145=350x^2+170x+20\)

\(\Rightarrow635x=125\Rightarrow x=\dfrac{25}{127}\)

Lập bảng thống kê giá trị.               

4x + 7 = 10x - 35

7 = 6x - 35

6x = 35 + 7

6x = 42

x = 42 : 6

x = 7

Vậy x = 7

# Học tốt #

4x+7=10x-35

7+35=10x-4x

42=6x

x=7

4x+7=10x-35

4x+7+35=10x

4x+42=10x

42=10x-4x

42=6x

x=42:6

x=7

TK CHO MK NHÉ

CHÚC BN HOK TỐT

a) \(\left|x-1\right|=2x-5\)

khi \(x\ge\frac{5}{2}\), phương trình có dạng:

\(\orbr{\begin{cases}x-1=2x-5\\x-1=5-2x\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=4\\x=2\end{cases}}\)x=4 thỏa mãn ĐK; x=2 không thỏa mãn ĐK

vậy phương trình có tập nghiệm là: \(S=\left\{4\right\}\)

a,|x-1|=2x-5 

1) x-1 = 2x-5 

x-2x = -5 + 1

-x = -4

=> x= 4

2) x-1 = -2x+5 

x+2x = 5 + 1 

3x = 6

x= 2

b,|3-8x|<19 

=> |3-8x| ={ 0;1;2;...18}

1) 3-8x = { 0;1;2;....;18}

=> 8x = { 3; 2; 1; ... ; -15}

x= { 3/8; 1/4; 1/8;.....; -15/8} 

2) 3-8x = { -1;-2;....;-18}

8x = { 4; 5;....; 21 }

x= { 1/2 ; 5/8 ;... ; 21/8}

a) \(4x\left(5x+2\right)-\left(10x-3\right)\left(2x+7\right)=133\)

\(=\left(20x^2+8x\right)-\left(20x^2+64x-21\right)=133\)

\(=20x^2+8x-20x^2-64x+21=133\)

\(=-56x=112\)

\(\Leftrightarrow x=-2\)

b) \(4\left(x-1\right)\left(x+5\right)+\left(x+2\right)\left(x+5\right)=5\left(x-1\right)\left(x+2\right)\)

\(\Leftrightarrow\left(x+5\right)\left(4x-4+x+2\right)=5\left(x-1\right)\left(x+2\right)\)

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

\(\Leftrightarrow5x^2+23x-10=5x^2+5x-10\)

\(\Leftrightarrow23x=5x\)

\(\Leftrightarrow18x=0\Leftrightarrow x=0\)

a) \(4x\left(5x+2\right)-\left(10x-3\right)\left(2x+7\right)=133\)

\(\Leftrightarrow20x^2+8x-\left[10x.\left(2x+7\right)-3.\left(2x+7\right)\right]=133\)

\(\Leftrightarrow20x^2+8x-\left(20x^2+70x-6x-21\right)=133\)

\(\Leftrightarrow20x^2+8x-20x^2-70x+6x+21=133\)

\(\Leftrightarrow8x-70x+6x=133-21\)

\(\Leftrightarrow-56x=112\)

\(\Leftrightarrow x=-\frac{112}{56}=-2\)

Vậy : \(x=-2\)