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\(\dfrac{x}{y}=\dfrac{3}{4}\Rightarrow\dfrac{x}{3}=\dfrac{y}{4}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{-3x}{-9}=\dfrac{5y}{20}=\dfrac{-3x+5y}{-9+20}=\dfrac{33}{11}=3\)
\(\Rightarrow\left\{{}\begin{matrix}x=3.3=9\\y=3.4=12\end{matrix}\right.\)
Đề trước đó:
(x-7)(x+1)-(x-3)^2=(3x-5)(3x+5)-(3x+1)^2+(x-2)^2-x
<=>x^2+x-7x-7-x^2+6x-9=9x^2-25-9x^2-6x-1+x^2-4x+4-x
<=>x^2-11x-6=0
<=>x^2-2x. 11/2 + 121/4-145/4=0
<=>(x-11/2)^2=145/4
<=>|x-11/2|=căn(145)/2
<=>x=[11+-căn(145)]/2
x+5 = 3x+1
x+4 = 3x (trừ 2 vế cho 1)
(x+4):x = 3x:x (chia 2 vế cho x)
x:x+4:x = 3
1+4:x = 3
4:x = 2 (trừ 2 vế cho 1)
=) x = 4:2 = 2
đúng nhé bạn
b) (5/2-3x)=25/9
3x = 5/2-25/9
3x =-5/18
x =-5/18:3
x=-5/54
\(e.\left(x-1\right)^5=-32\)
\(\left(x-1\right)^5=\left(-2\right)^5\)
\(x-1=-2\)
\(x\) \(=-2+1\)
\(x\) \(=-1\)
Vậy \(x=-1\)
\(\left|x+1\right|+\left|x-5\right|=3x+1\left(đk:x\ge-\dfrac{1}{3}\right)\)
\(\Leftrightarrow x+1+\left|x-5\right|=3x+1\)
\(\Leftrightarrow\left|x-5\right|=2x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=2x\left(x\ge5\right)\\x-5=-2x\left(-\dfrac{1}{3}\le x< 5\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\left(ktm\right)\\x=\dfrac{5}{3}\left(tm\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-x-1+5-x=3x+1\left(x< -1\right)\\x+1+5-x=3x+1\left(-1\le x< 5\right)\\x+1+x-5=3x+1\left(x\ge5\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\left(ktm\right)\\x=\dfrac{5}{3}\left(tm\right)\\x=-5\left(ktm\right)\end{matrix}\right.\Rightarrow x=\dfrac{5}{3}\)