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a) \(\frac{11}{10,5}=\frac{6,32}{-x}\)
\(6,32:\frac{11}{10,5}=x\)
\(\frac{1659}{275}=x\)
b) \(\frac{x-1}{x+5}=\frac{6}{7}\)
\(\Rightarrow7x-7=6x+30\)
\(7x-6x=30+7\)
\(x=37\)
c) \(\frac{x^2}{6}=\frac{24}{25}\)
\(x^2=\frac{144}{25}\)
\(x=\frac{12}{5}\)
d) \(x:0,16=9:x\)
\(\frac{x}{0,16}=\frac{9}{x}\)
\(x^2=1,44\)
\(x=1,2\)
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\(\sqrt{0,16}\)-\(\sqrt{\dfrac{1}{25}}\)=\(\dfrac{2}{5}\)-\(\dfrac{1}{5}\)=\(\dfrac{1}{5}\)
Dễ mà sao mà ko biết làm hay dậy!
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+) Có: \(x:y:z:t=2:3:4:5\)
\(\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{t}{5}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{t}{5}=\frac{x+y+z+t}{2+3+4+5}=\frac{-42}{14}=-3\\ \Rightarrow\left\{{}\begin{matrix}\frac{x}{2}=-3\Rightarrow x=\left(-3\right)\cdot2=-6\\\frac{y}{3}=-3\Rightarrow y=\left(-3\right)\cdot3=-9\\\frac{z}{4}=-3\Rightarrow z=\left(-3\right)\cdot4=-12\\\frac{t}{5}=-3\Rightarrow t=\left(-3\right)\cdot5=-15\end{matrix}\right.\)
Vậy \(x=-6;y=-9;z=-12;t=-15\)
+) Gọi giá trị chung của tỉ lệ thức là k, ta có:
\(\frac{x}{4}=\frac{y}{7}=k\\ \Rightarrow x=4k;y=7k\)
Lại có: \(x\cdot y=112\)
\(\Rightarrow4k\cdot7k=112\\ 28k^2=112\\ \Rightarrow k^2=4\\ \Rightarrow k=\pm2\)
\(\Rightarrow\left\{{}\begin{matrix}x=4k=4\cdot\left(\pm2\right)=\pm8\\y=7k=7\cdot\left(\pm2\right)=\pm14\end{matrix}\right.\)
Vậy \(x=\pm8;y=\pm14\)
+) Gọi giá trị chung của tỉ lệ thức là h, ta có:
\(\frac{x}{3}=\frac{y}{4}=h\\ \Rightarrow x=3h;y=4h\)
Lại có: \(x\cdot y=48\)
\(\Rightarrow3h\cdot4h=48\\ 12h^2=48\\ \Rightarrow h^2=4\\ \Rightarrow h=\pm2\)
\(\Rightarrow\left\{{}\begin{matrix}x=3h=3\cdot\left(\pm2\right)=\pm6\\y=4h=4\cdot\left(\pm2\right)=\pm8\end{matrix}\right.\)
Vậy \(x=\pm6;y=\pm8\)
+) Gọi giá trị chung của tỉ lệ thức là g, ta có:
\(\frac{x}{2}=\frac{y}{-3}=g\\ \Rightarrow x=2g;y=-3g\)
Mà \(xy=-54\)
\(\Rightarrow2g\cdot\left(-3g\right)=-54\\ -6g^2=-54\\ g^2=9\\ \Rightarrow g=\pm3\)
\(\Rightarrow\left\{{}\begin{matrix}x=2g=2\cdot\left(\pm3\right)=\pm6\\y=-3g=\left(-3\right)\cdot\left(\pm3\right)=\pm9\end{matrix}\right.\)
Vậy \(x=\pm6;y=\pm9\)
+) \(\left(x-2\right)^{2012}+\left|y^2-9\right|^{2014}=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x-2\right)^{2012}=0\\\left|y^2-9\right|^{2014}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-2=0\\\left|y^2-9\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y^2-9=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y^2=9\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y=\pm3\end{matrix}\right.\)
Vậy \(x=2;y=\pm3\)
+) \(-0,16:x=-x:25\)
\(-0,16\cdot25=-x\cdot x\\ -x^2=-4\\ \Rightarrow x^2=4\\ \Rightarrow x=\pm2\)
Vậy \(x=\pm2\)
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\(\frac{4\left(3x-2y\right)}{16}=\frac{3\left(2z-4x\right)}{9}=\frac{2\left(4y-3z\right)}{4}=\frac{12x-8y+6z-12x+8y-6z}{16+9+4}=\frac{0}{29}=0\)
\(\Leftrightarrow3x-2y=0\Leftrightarrow\frac{x}{2}=\frac{y}{3}\)
\(\Leftrightarrow2z-4x=0\Leftrightarrow\frac{x}{2}=\frac{z}{4}\)
\(\Leftrightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)
\(\Leftrightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{2x+4y+5z}{4+12+20}=\frac{8}{36}=\frac{2}{9}=\frac{2x+3y-z}{4+12-4}\)=> A= 2x+3y -z = 12.2/9 =8/3
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0,16:x=x:36
0,16/x=x/36
x^2=144/25
x^2=(12/5)^2;x^2=-(12/5)^2
x=12/5. ; x=-12/5
tick nha bn
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c)/x+\(\frac{4}{5}\)/=\(\frac{1}{7}\)
=>x+\(\frac{4}{5}\)=\(\frac{1}{7}\) hoặc x+\(\frac{4}{5}\)=\(\frac{-1}{7}\)
x=\(\frac{1}{7}\)-\(\frac{4}{5}\) x=\(\frac{-1}{7}\)-\(\frac{4}{5}\)
x=\(\frac{-23}{35}\) x=\(\frac{-33}{35}\)
a)x:\(\left(\frac{-1}{3}\right)^3\)=\(\left(\frac{-1}{3}\right)^1\)
x=\(\left(\frac{-1}{3}\right)^1\)*\(\left(\frac{1}{-3}\right)^3\)
x=\(\left(\frac{-1}{3}\right)^4\)
x=\(\frac{1}{81}\)
b)x:0,16=9:x
=>x:0,16:x=9:x:x
=>0,16=9:\(x^2\)
=>9:\(x^2\)=0,16
=>\(x^2\)=9:0,16
=>\(x^2\)=\(\frac{225}{4}\)
=>\(x\)=\(\frac{15}{2}\)
\(\frac{0,16}{x}=\frac{x}{25}\)
\(x.x=0,16.25\)
\(x.x=\left(0.4\right)^2.5^2\)
\(x=2;x=-2\)
Biết x= 2 hoặc -2 rồi nhưng mà cách làm ...