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phạm bích ngọc
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Nguyễn Minh Tuyền
8 tháng 7 2018 lúc 12:03

\(\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow8x+16-5x^2-10x+4x^2+4x-8x-8+2x^2-8=0\)

\(\Leftrightarrow x^2-6x=0\Leftrightarrow x\left(x-6\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x-6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}}\)

                                    Vậy S = { 0, 6}

super xity
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Trần Tuyết Như
19 tháng 8 2015 lúc 14:37

 (8 - 5x) (x + 2) + 4(x - 2) (x + 1) + 2(x - 2) (x + 2) = 0

=>  (x + 2) [ (8 - 5x) + 4(x + 1) + 2(x - 2)] = 0

=> (x + 2) (8 - 5x + 4x + 4 + 2x - 4)  = 0

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

=> x + 2 = 0   hoặc      x + 8 = 0

=> x = -2       hoặc        x = -8

Lâm Hữu
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super xity
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lemailinh
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|5\(x\) - 4| = |\(x+2\)|

\(\left[{}\begin{matrix}5x-4=x+2\\5x-4=-x-2\end{matrix}\right.\)

\(\left[{}\begin{matrix}4x=6\\6x=2\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

vậy \(x\in\) { \(\dfrac{1}{3};\dfrac{3}{2}\)}

|2\(x\) - 3| - |3\(x\) + 2| = 0

|2\(x\) - 3| = | 3\(x\) + 2|

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

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

vậy \(x\in\){ -5; \(\dfrac{1}{5}\)}

 

 

|\(\dfrac{5}{4}\)\(x\) - \(\dfrac{7}{2}\)| - | \(\dfrac{5}{8}\)\(x\) + \(\dfrac{3}{5}\)| = 0

|\(\dfrac{5}{4}x\) - \(\dfrac{7}{2}\)|    = | \(\dfrac{5}{8}x+\dfrac{3}{5}\)|

\(\left[{}\begin{matrix}\dfrac{5}{4}x-\dfrac{7}{2}=\dfrac{5}{8}x+\dfrac{3}{5}\\\dfrac{5}{4}x-\dfrac{7}{2}=-\dfrac{5}{8}x-\dfrac{3}{5}\end{matrix}\right.\)

  \(\left[{}\begin{matrix}\dfrac{5}{4}x-\dfrac{5}{8}x=\dfrac{3}{5}+\dfrac{7}{2}\\\dfrac{5}{4}x+\dfrac{5}{2}x=-\dfrac{3}{5}+\dfrac{7}{2}\end{matrix}\right.\)

 \(\left[{}\begin{matrix}\dfrac{5}{8}x=\dfrac{41}{10}\\\dfrac{15}{8}x=\dfrac{29}{10}\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{164}{25}\\x=\dfrac{116}{75}\end{matrix}\right.\)

Vậy \(x\in\) { \(\dfrac{116}{75}\)\(\dfrac{164}{25}\)}

Nguyễn Thị Hồng Duyên
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Hiếu Tuấn
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Nguyễn Lê Phước Thịnh
5 tháng 11 2023 lúc 19:56

2: \(3x\left(x-4\right)+2x-8=0\)

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

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

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

3: 4x(x-3)+x2-9=0

=>\(4x\left(x-3\right)+\left(x+3\right)\left(x-3\right)=0\)

=>\(\left(x-3\right)\left(4x+x+3\right)=0\)

=>\(\left(x-3\right)\left(5x+3\right)=0\)

=>\(\left[{}\begin{matrix}x-3=0\\5x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{5}\end{matrix}\right.\)

4: \(x\left(x-1\right)-x^2+3x=0\)

=>\(x^2-x-x^2+3x=0\)

=>2x=0

=>x=0

5: \(x\left(2x-1\right)-2x^2+5x=16\)

=>\(2x^2-x-2x^2+5x=16\)

=>4x=16

=>x=4

Huy Anh
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Huỳnh Quang Sang
9 tháng 10 2020 lúc 20:53

(8 - 5x)(x + 2) + 4(x - 2)(x + 1) + 2(x - 2)(x + 2) = 0

=> 8(x + 2) - 5x(x + 2) + 4[x(x + 1) - 2(x + 1)] + 2(x2 - 4) = 0

=> 8x + 16 - 5x2 - 10x + 4(x2 + x - 2x - 2) + 2x2 - 8 = 0

=> 8x + 16 - 5x2 - 10x + 4x2 + 4x - 8x - 8 + 2x2 - 8 = 0

=> (8x - 10x + 4x - 8x) + (16 - 8 - 8) + (-5x2 + 4x2 + 2x2)  = 0

=> 0 + x2 = 0

=> x2 = 0 => x = 0

Khách vãng lai đã xóa
Greninja
9 tháng 10 2020 lúc 20:55

\(\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)=0\)

\(-5x^2-2x+16+4\left(x^2-x-2\right)+2\left(x^2-4\right)=0\)

\(-5x^2-2x+16+4x^2-4x-8+2x^2-8=0\)

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

\(x\left(x-6\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x-6=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=6\end{cases}}\)

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l҉o҉n҉g҉ d҉z҉
9 tháng 10 2020 lúc 20:56

( 8 - 5x )( x + 2 ) + 4( x - 2 )( x + 1 ) + 2( x - 2 )( x + 2 ) = 0

<=> ( x + 2 )[ ( 8 - 5x ) + 2( x - 2 ) ] + 4( x2 - x - 2 ) = 0

<=> ( x + 2 )( 8 - 5x + 2x - 4 ) + 4x2 - 4x - 8 = 0

<=> ( x + 2 )( 4 - 3x ) + 4x2 - 4x - 8 = 0

<=> 4x - 3x2 + 8 - 6x + 4x2 - 4x - 8 = 0

<=> x2 - 6x = 0

<=> x( x - 6 ) = 0

<=> \(\orbr{\begin{cases}x=0\\x-6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}\)

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Haly
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HT.Phong (9A5)
19 tháng 6 2023 lúc 14:42

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

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

\(\Rightarrow x=0\)

\(\left(x+3\right)+1=7\)

\(\Rightarrow x+3+1=7\)

\(\Rightarrow x+4=7\)

\(\Rightarrow x=3\)

\(\left(3x-4\right)+4=12\)
\(\Rightarrow3x-4+4=12\)

\(\Rightarrow3x=12\)

\(\Rightarrow x=4\)

\(\left(5x+4\right)-1=13\)

\(\Rightarrow5x+4-1=13\)

\(\Rightarrow5x+3=13\)

\(\Rightarrow5x=10\)

\(\Rightarrow x=2\)

\(\left(4x-8\right)-3=5\)

\(\Rightarrow4x-8-3=5\)

\(\Rightarrow4x-11=5\)

\(\Rightarrow4x=16\)

\(\Rightarrow x=4\)

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

\(\Rightarrow8-2x-4=2\)

\(\Rightarrow4-2x=2\)

\(\Rightarrow2x=2\)

\(\Rightarrow x=1\)

\(7+\left(5x+2\right)=14\)

\(\Rightarrow7+5x+2=14\)

\(\Rightarrow9+5x=14\)

\(\Rightarrow5x=5\)

\(\Rightarrow x=1\)

\(5-\left(3x-11\right)=1\)

\(\Rightarrow5-3x+11=1\)

\(\Rightarrow16-3x=1\)

\(\Rightarrow3x=15\)

\(\Rightarrow x=5\)