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biết giải bài 2
x/12=y/14=x.y/12.24=98/288=49/144
=> x/12=49/144=> 49/12
=> y/14=49/144=> 343/72
mới lớp 2 thôi
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a) Ta có : (3x - 0.5) ( 2x + 2.5) = 0
\(\Leftrightarrow\orbr{\begin{cases}3x-0,5=0\\2x+2,5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x=0,5\\2x=-2,5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{0,5}{3}=\frac{1}{6}\\x=-\frac{2,5}{2}=\frac{5}{4}\end{cases}}\)
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d. \(\left|x-\frac{2}{5}\right|+\frac{1}{2}=\frac{3}{4}\)
\(\left|x-\frac{2}{5}\right|=\frac{3}{4}-\frac{1}{2}\)
\(\left|x-\frac{2}{5}\right|=\frac{3-2}{4}\)
\(\left|x-\frac{2}{5}\right|=\frac{1}{4}\)
Vậy \(x-\frac{2}{5}=\frac{1}{4}\) hoặc \(x-\frac{2}{5}=\frac{-1}{4}\)
* \(x-\frac{2}{5}=\frac{1}{4}\)
\(x=\frac{1}{4}+\frac{2}{5}\)
\(x=\frac{5+8}{20}\)
\(x=\frac{13}{20}\)
* \(x-\frac{2}{5}=\frac{-1}{4}\)
\(x=\frac{-1}{4}+\frac{2}{5}\)
\(x=\frac{-5+8}{20}\)
\(x=\frac{3}{20}\)
Vậy \(x\in\left\{\frac{13}{20};\frac{3}{20}\right\}\)
d) |x-2/5| +1/2 = 3/4
=> |x-2/5| = 1/4
<=> x-2/5 = 1/4 hoặc x=2/5 = -1/4
<=> x=13/20 hoặc 3/20
e) 5/6 -| 2-x| = 1/3
=> |2-x|=1/2
<=> 2-x=1/2 hoặc 2-x=-1/2
<=> x=3/2 hoặc x=5/2
g) làm tương tự
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x^2-3x+3=0\)
\(\Leftrightarrow x^2-2.x.\frac{3}{2}+\left(\frac{3}{2}\right)^2+\frac{3}{4}=0\)
\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2+\frac{3}{4}=0\)
\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2=\frac{-3}{4}\)
Vô lí => Phương trình vô nghiệm
\(\Rightarrow x\in\varnothing\)
b) \(x-\left(x-2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow x-x+2-x+2=0\)
\(\Leftrightarrow-x=-4\)
\(\Leftrightarrow x=4\)
c) \(\left(x-2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=1\end{cases}}}\)
Vậy x = {1;2}
d) \(x^2-2x-x+2=0\)
\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}}\)
Vậy x = {1;2}
![](https://rs.olm.vn/images/avt/0.png?1311)
2.(x-1)-3.(2x+2)-4.(2x+3)=16
=>2x-2-6x-6-8x-12=16
=>2x-6x-8x-(2+6+12)=16
=>x.(2-6-8)=16+20=36
=>x.(-12)=36
=>x=-3
Vậy x=-3
\(2\left(x-1\right)-3\left(2x+2\right)-4\left(2x+3\right)=16\)
\(\Leftrightarrow2x-2-6x-6-8x-12=16\)
\(\Leftrightarrow\left(2x-6x-8x\right)+\left(-2-6-12\right)=16\)
\(\Leftrightarrow-12x-20=16\)
\(\Leftrightarrow-12x=36\)
\(\Leftrightarrow x=\frac{-36}{12}-3\)
\(\left|x-1\right|+\left|x-4\right|=3x\).
Xét \(x< 1\)thì \(\left|x-1\right|=1-x\); \(\left|x-4\right|=4-x\) . Do đó:
\(\left(1-x\right)+\left(4-x\right)=3x\).
\(\Rightarrow5-2x=3x\).
\(\Rightarrow5x=5\).
\(\Rightarrow x=1\)(không thuộc khoảng đang xét).
Xét \(1\le x\le4\)thì \(\left|x-1\right|=x-1\); \(\left|x-4\right|=4-x\). Do đó:
\(\left(x-1\right)+\left(4-x\right)=3x\).
\(\Rightarrow3=3x\).
\(\Rightarrow x=1\)(thuộc khoảng đang xét).
Xét \(x>4\)thì \(\left|x-1\right|=x-1\); \(\left|x-4\right|=x-4\). Do đó:
\(\left(x-1\right)+\left(x-4\right)=3x\).
\(\Rightarrow2x-5=3x\).
\(\Rightarrow x=-5\)(không thuộc khoảng đang xét).
Vậy \(x=1\).
\(\left|a\right|=a\)nếu \(a\ge0\).
\(\left|a\right|=-a\)nếu \(a\le0\).