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\(\dfrac{1}{2}-3x+\left|x-1\right|=0\\ \Rightarrow3x+\left|x-1\right|=\dfrac{1}{2}-0\\ \Rightarrow3x+\left|x-1\right|=\dfrac{1}{2}\\ \Rightarrow\left|x-1\right|=\dfrac{1}{2}-3x\\ \Rightarrow\left[{}\begin{matrix}x-1=\dfrac{1}{2}-3x\\x-1=-\dfrac{1}{2}+3x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x+3x=\dfrac{1}{2}+1\\x-3x=-\dfrac{1}{2}+1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}4x=\dfrac{3}{2}\\2x=\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{8}\\x=\dfrac{1}{4}\end{matrix}\right.\)
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\(\dfrac{1}{2}\left|2x-1\right|+\left|2x-1\right|=x+1\\ \Rightarrow\left|2x-1\right|\cdot\left(\dfrac{1}{2}+1\right)=x+1\\ \Rightarrow\left|2x-1\right|\cdot\dfrac{3}{2}=x+1\\ \Rightarrow\left|2x-1\right|=x+1:\dfrac{3}{2}\\ \Rightarrow\left|2x-1\right|=x+\dfrac{2}{3}\\ \Rightarrow\left[{}\begin{matrix}2x-1=x+\dfrac{2}{3}\\2x-1=-x-\dfrac{2}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-x=\dfrac{2}{3}+1\\2x+x=-\dfrac{2}{3}+1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\3x=\dfrac{1}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{1}{9}\end{matrix}\right.\)
(3x+5)(2x-7)=0
\(\Leftrightarrow\left[{}\begin{matrix}3x+5=0\\2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-5\\2x=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-5}{3}\\x=\frac{7}{2}\end{matrix}\right.\)
(-5x+2)(-3x-4)=0
\(\Leftrightarrow\left[{}\begin{matrix}\left(-5x+2\right)=0\\\left(-3x-4\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-5x=-2\\-3x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{5}\\x=\frac{-3}{4}\end{matrix}\right.\)
(x-5)(4x-3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\4x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\frac{3}{4}\end{matrix}\right.\)
-2x(x+1)(x-1)=0
\(\Leftrightarrow\left[{}\begin{matrix}-2x=0\\x+1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=1\end{matrix}\right.\)
\(\left(3x+5\right).\left(2x-7\right)=0\)
=> \(\left\{{}\begin{matrix}3x+5=0\\2x-7=0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}3x=0-5=-5\\2x=0+7=7\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=\left(-5\right):3\\x=7:2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=-\frac{5}{3}\\x=\frac{7}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{5}{3};\frac{7}{2}\right\}\).
\(\left(-5x+2\right).\left(-3x-4\right)=0\)
=> \(\left\{{}\begin{matrix}-5x+2=0\\-3x-4=0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}-5x=0-2=-2\\-3x=0+4=4\end{matrix}\right.\) =>\(\left\{{}\begin{matrix}x=\left(-2\right):\left(-5\right)\\x=4:\left(-3\right)\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=\frac{2}{5}\\x=-\frac{4}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{2}{5};-\frac{4}{3}\right\}\).
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mi tích tau tau tích mi xong tau trả lời nka
việt nam nói là làm
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) 5/4+x=2/3
B) -x-2=5/4
C)4x+1/3=3/2
Đ) 1/3-2/5+3x=3/4
E) 3x+7+2x=4x-3
G) 3x(2x-3)-2x(3x-4)=15
H) x^2-x=0
a) \(x=-\frac{7}{12}\)
b) \(x=-\frac{13}{4}\)
c) \(x=\frac{7}{24}\)
d) \(x=\frac{49}{180}\)
e) \(x=-10\)
g) \(x=15\)
h) \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
1) ( 5 - 3x ) ( 1 + x ) = 0
=> \(\orbr{\begin{cases}5-3x=0\\1+x=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}\)
Vậy x = - 1 hoặc x = \(\frac{5}{3}\)
2) x2 - 3x = 0
=> x ( x - 3 ) = 0
=> \(\orbr{\begin{cases}x=0\\x-3=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
Vậy x = 0 hoặc x = 3
3) 4x + 2x2 = 0
=> 2x ( x + 2 ) = 0
=> \(\orbr{\begin{cases}2x=0\\x+2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)
Vậy x = 0 hoặc x = - 2
\(a,\left(5-3x\right)\left(1+x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5-3x=0\\1+x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=5\\x=-1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}}\)
vậy_
\(b,x^2-3x=0\)
\(\Leftrightarrow x\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}}\)
vậy_
\(c,4x+2x^2=0\)
\(\Leftrightarrow x\left(4+2x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\4+2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\2x=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=-2\end{cases}}}\)
vậy_