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a, \(\frac{\left(5-2x\right)}{3}=\frac{\left(4x-1\right)}{-5}\)
\(\Leftrightarrow-5(5-2x)=3\left(4x-1\right)\)
\(\Leftrightarrow10x-25=12x-3\)
\(\Leftrightarrow10x-12x=25-3\)
\(\Leftrightarrow-2x=22\)
\(\Leftrightarrow x=-11\)
b, \(\frac{\left(12-3x\right)}{32}=\frac{6}{\left(4-x\right)}\)
\(\Leftrightarrow\frac{3\left(4-x\right)}{32}=\frac{6}{\left(4-x\right)}\)
\(\Leftrightarrow3(4-x)\left(4-x\right)=32.6\)
\(\Leftrightarrow(4-x)\left(4-x\right)=32.2\)
\(\Leftrightarrow(4-x)^2=64\)
\(\Leftrightarrow\orbr{\begin{cases}4-x=8\\4-x=-8\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-4\\x=12\end{cases}}\)
c, \(\frac{\left(10-2x\right)}{6}=\frac{27}{\left(5-x\right)}\)
\(\Leftrightarrow\frac{2\left(5-x\right)}{6}=\frac{27}{\left(5-x\right)}\)
\(\Leftrightarrow2(5-x)\left(5-x\right)=27.6\)
\(\Leftrightarrow(5-x)\left(5-x\right)=27.3\)
\(\Leftrightarrow(5-x)^2=81\)
\(\Leftrightarrow\orbr{\begin{cases}5-x=9\\5-x=-9\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-4\\x=14\end{cases}}\)
a, \(\frac{5-2x}{3}=\frac{4x-1}{-5}\Leftrightarrow-25+10x=12x-3\Leftrightarrow-22-2x=0\Leftrightarrow x=-11\)
b, \(\frac{12-3x}{32}=\frac{6}{4-x}\Leftrightarrow\frac{12-3x}{32}=\frac{18}{12-3x}\)
\(\Leftrightarrow\left(12-3x\right)^2=576\Leftrightarrow12-3x=\pm2\)\(\Leftrightarrow\orbr{\begin{cases}x=\frac{10}{3}\\x=\frac{14}{3}\end{cases}}\)
c, \(\frac{10-2x}{6}=\frac{27}{5-x}\Leftrightarrow\frac{10-2x}{6}=\frac{54}{10-2x}\)
\(\Leftrightarrow\left(10-2x\right)^2=324\Leftrightarrow10-2x=\pm18\)\(\Leftrightarrow\orbr{\begin{cases}x=14\\x=-4\end{cases}}\)
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1) \(4x=7y\Leftrightarrow\dfrac{x}{7}=\dfrac{y}{4}\Rightarrow\dfrac{x^2}{49}=\dfrac{y^2}{16}=\dfrac{x^2+y^2}{49+16}=\dfrac{260}{65}=4\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2=4.49=196\\y^2=4.16=64\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=14,y=8\\x=-14,y=-8\end{matrix}\right.\) (vì \(\dfrac{x}{7}=\dfrac{y}{4}\) nên \(x,y\) cùng dấu)
2) \(2^{x-1}+5.2^{x-2}=\dfrac{7}{32}\)
\(\Leftrightarrow2^{x-1}+\dfrac{5}{2}.2^{x-1}=\dfrac{7}{32}\)
\(\Leftrightarrow2^{x-1}=\dfrac{1}{16}=2^{-4}\)
\(\Leftrightarrow x-1=-4\)
\(\Leftrightarrow x=-3\)
3) \(\left|x+5\right|+\left(3y-4\right)^{2016}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+5=0\\3y-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-5\\y=\dfrac{4}{3}\end{matrix}\right.\)
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Các đa thức 1 biến là : A, B, M, N là những đa thức một biến.
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a: Ta có: 2x/3=3y/4=4z/5
nên \(\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}\)
Đặt \(\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}=k\)
=>x=3/2k; y=4/3k; z=5/4k
\(xy+yz-xz=32\)
\(\Leftrightarrow\dfrac{3}{2}k\cdot\dfrac{4}{3}k+\dfrac{4}{3}k\cdot\dfrac{5}{4}k-\dfrac{3}{2}k\cdot\dfrac{5}{4}k=32\)
\(\Leftrightarrow k^2\cdot\dfrac{43}{24}=32\)
\(\Leftrightarrow k^2=\dfrac{768}{43}\)
Trường hợp 1: \(k=\dfrac{16\sqrt{129}}{43}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{24\sqrt{129}}{43}\\y=\dfrac{64\sqrt{129}}{129}\\z=\dfrac{20\sqrt{129}}{43}\end{matrix}\right.\)
Trường hợp 2: \(k=-\dfrac{16\sqrt{129}}{43}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{24\sqrt{129}}{43}\\y=-\dfrac{64\sqrt{129}}{129}\\z=-\dfrac{20\sqrt{129}}{43}\end{matrix}\right.\)
b: Ta có: 4x=3y
nên x/3=y/4=k
=>x=3k; y=4k
\(x^2-xy+y^2=32\)
\(\Leftrightarrow9k^2-12k^2+16k^2=32\)
\(\Leftrightarrow13k^2=32\)
Trường hợp 1: \(k=\dfrac{32\sqrt{13}}{13}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{96\sqrt{13}}{13}\\y=\dfrac{128\sqrt{13}}{13}\end{matrix}\right.\)
Trường hợp 2: \(k=-\dfrac{32\sqrt{13}}{13}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{96\sqrt{13}}{13}\\y=-\dfrac{128\sqrt{13}}{13}\end{matrix}\right.\)
( 4x - 1)^5=32
=> x= 0,75
đấp án là 0,75
nhớ cho mik nha( cần giúp thì liên hệ và kết bạn với tớ tớ sẽ giúp gỡ các bài toán phức tạp từ 1-12)
(4x-1)5=32
(4x-1)5=25
4x-1=2
4x=3
x=0,75