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bài 1)
a) \(\dfrac{11}{13}-\left(\dfrac{5}{42}-x\right)=-\left(\dfrac{15}{28}-\dfrac{11}{15}\right)
\)
\(\left(\dfrac{5}{42}-x\right)=\dfrac{11}{13}+\dfrac{15}{28}-\dfrac{11}{15}\)
\(x=\dfrac{5}{42}-\dfrac{3541}{5460}=-\dfrac{413}{780}\)
b) \(\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|2,15\right|\)
\(\left|x+\dfrac{4}{15}\right|=-\left|2,15\right|+\left|3,75\right|=1,6\)
\(\Rightarrow x+\dfrac{4}{15}=1,6\) hoặc \(x+\dfrac{4}{15}=-1,6\)
\(\Rightarrow x=\dfrac{4}{3}\) hoặc \(x=-\dfrac{28}{15}\)
c) \(\dfrac{5}{3}-\left|x-\dfrac{3}{2}\right|=-\dfrac{1}{2}\)
\(\Rightarrow\left|x-\dfrac{3}{2}\right|=\dfrac{5}{3}+\dfrac{1}{2}=\dfrac{13}{6}\)
\(\Rightarrow x-\dfrac{3}{2}=\dfrac{13}{6}\) hoặc \(x-\dfrac{3}{2}=-\dfrac{13}{6}\)
\(\Rightarrow x=\dfrac{11}{3}\) hoặc \(x=-\dfrac{2}{3}\)
d)\(\left(x-\dfrac{2}{3}\right).\left(2x-\dfrac{3}{2}\right)=0\)
\(\Rightarrow x-\dfrac{2}{3}=0\) hoặc \(2x-\dfrac{3}{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{3}{4}\end{matrix}\right.\)
3) a) \(\left(x^{^2}-4\right)^{^2}+\left(x+2\right)^{^2}=0\)
Vì \(\left(x^{^2}-4\right)^{^2}\ge0,\left(x+2\right)^{^2}\ge0\) nên :
\(\left\{{}\begin{matrix}x^{^2}-4=0\\x+2=0\end{matrix}\right.\Rightarrow x=\pm2\)
b) \(\left(x-y\right)^{^2}+\left|y+2\right|=0\)
Vì \(\left\{{}\begin{matrix}\left(x-y\right)^{^2}\ge0\\\left|y+2\right|\ge0\end{matrix}\right.\) nên \(\left\{{}\begin{matrix}x-y=0\\y+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-y=0\\y=-2\end{matrix}\right.\Rightarrow x=-2;y=-2\)
c) \(\left|x-y\right|+\left|y+\dfrac{9}{25}\right|=0\)
Vì \(\left\{{}\begin{matrix}\left|x-y\right|\ge0\\\left|y+\dfrac{9}{25}\right|\ge0\end{matrix}\right.\) nên \(\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{25}=0\end{matrix}\right.\Rightarrow y=-\dfrac{9}{25};x=-\dfrac{9}{25}\)
d) \(\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|=\left(-\dfrac{1}{4}\right)-\left|y\right|\)
\(\Rightarrow\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|+\left|y\right|=-\dfrac{1}{4}\)
Vì \(\left\{{}\begin{matrix}\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|\ge0\\\left|y\right|\ge0\end{matrix}\right.\) mà \(\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|+\left|y\right|=-\dfrac{1}{4}\) nên không tồn tại x,y thỏa mãn đề bài .
b: 2x^3-1=15
=>2x^3=16
=>x=2
\(\dfrac{x+16}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}\)
=>\(\dfrac{y-25}{16}=\dfrac{z+9}{25}=\dfrac{18}{9}=2\)
=>y-25=32; z+9=50
=>y=57; z=41
d: 3/5x=2/3y
=>9x=10y
=>x/10=y/9=k
=>x=10k; y=9k
x^2-y^2=38
=>100k^2-81k^2=38
=>19k^2=38
=>k^2=2
TH1: k=căn 2
=>\(x=10\sqrt{2};y=9\sqrt{2}\)
TH2: k=-căn 2
=>\(x=-10\sqrt{2};y=-9\sqrt{2}\)
Bài 1:
a: \(\Leftrightarrow\dfrac{x+2}{2}=x-5\)
=>2x-10=x+2
=>x=12
b: \(\Leftrightarrow\left(x+2\right)^2=100\)
=>x+2=10 hoặc x+2=-10
=>x=-12 hoặc x=8
c: \(\Leftrightarrow\left(2x-5\right)^3=27\)
=>2x-5=3
=>2x=8
=>x=4
a, \(\left[x\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\right]x^2-1\)
\(=\left[x\left(x^2-16\right)-\left(x^2+1\right)\right]x^2-1\)
\(=\left[x^3-16x-x^2-1\right]x^2-1\)
\(=x^5-16x^3-x^4-x^2-1\)
b, \(\left(y-3\right)y+3y^2+9-y^2+2\left(y^2-2\right)\)
\(=y^2-3y+3y^2+9-y^2+2y^2-4\)
\(=5y^2-3y+5\)
c, \(\left(x+y\right)\left(x^2x^2-xy+y^2\right)\)
\(=x^5-x^2y+xy^2+x^4y-xy^2+y^3\)
d, \(\left(\dfrac{1}{2}xy+\dfrac{3}{4}y\right).\dfrac{1}{2}xy-\dfrac{3}{4}y\)
\(=\dfrac{1}{4}x^2y^2+\dfrac{3}{8}xy^2-\dfrac{3}{4}y\)
\(=\dfrac{1}{4}y.\left(x^2y+\dfrac{3}{2}xy-3\right)\)
Chúc bạn học tốt!!!
a: \(\left(-\dfrac{3}{4}\right)^{3x-1}=\dfrac{256}{81}\)
\(\Leftrightarrow\left(-\dfrac{3}{4}\right)^{3x-1}=\left(-\dfrac{3}{4}\right)^{-4}\)
=>3x-1=-4
=>3x=-3
hay x=-1
b: \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
\(\Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-7=-1\\x-7=1\end{matrix}\right.\Leftrightarrow x\in\left\{7;6;8\right\}\)
c: \(\left(x-\dfrac{1}{2}\right)^2+\left(y+\dfrac{1}{2}\right)^2=0\)
=>x-1/2=0 và y+1/2=0
=>x=1/2 và y=-1/2
\(\dfrac{x-2}{4}=\dfrac{y+1}{5}=\dfrac{z+3}{7}\)
\(\Rightarrow\dfrac{2\left(x-2\right)}{8}=\dfrac{y+1}{5}=\dfrac{2\left(z+3\right)}{14}\)
\(\Rightarrow\dfrac{2x-4}{8}=\dfrac{y+1}{5}=\dfrac{2z+6}{14}\)
Dựa vào tính chất dãy tỉ số bằng nhau ta có:
\(=\dfrac{2x-4+y+1-2z-6}{8+5-14}\)
\(=\dfrac{2x+y-2z-9}{-1}\)
\(=\dfrac{7-9}{-1}=2\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-2}{4}=2\Rightarrow x-2=8\Rightarrow x=10\\\dfrac{y+1}{5}=2\Rightarrow y+1=10\Rightarrow y=9\\\dfrac{z+3}{7}=2\Rightarrow z+3=14\Rightarrow z=11\end{matrix}\right.\)
Giải:
a) \(x+\left(-\dfrac{31}{12}\right)^2=\left(\dfrac{49}{12}\right)^2-x=y\)
\(\Leftrightarrow x+\left(-\dfrac{31}{12}\right)^2=\left(\dfrac{49}{12}\right)^2-x\)
\(\Leftrightarrow x+\left(-\dfrac{31}{12}\right)^2-\left(\dfrac{49}{12}\right)^2+x=0\)
\(\Leftrightarrow2x+\left(-\dfrac{31}{12}\right)^2-\left(\dfrac{49}{12}\right)^2=0\)
\(\Leftrightarrow2x+\dfrac{\left(-31\right)^2}{12^2}-\dfrac{49^2}{12^2}=0\)
\(\Leftrightarrow2x+\dfrac{\left(-31\right)^2-49^2}{144}=0\)
\(\Leftrightarrow2x+\dfrac{961-2401}{144}=0\)
\(\Leftrightarrow2x+\dfrac{-1440}{144}=0\)
\(\Leftrightarrow2x+\left(-10\right)=0\)
\(\Leftrightarrow2x=10\)
\(\Leftrightarrow x=5\)
Mà \(x+\left(-\dfrac{31}{12}\right)^2=y^2\)
\(\Leftrightarrow5+\dfrac{961}{144}=y^2\)
\(\Leftrightarrow y^2=\dfrac{1681}{144}\)
\(\Leftrightarrow y=\pm\dfrac{41}{12}\)
Vậy ...
b) \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\)
Vì \(\left(\dfrac{1}{2}x-5\right)^{20}\ge0;\forall x\)
và \(\left(y^2-\dfrac{1}{4}\right)^{10}\ge0;\forall y\)
\(\Rightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\)
\(\Leftrightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x-5=0\\y^2-\dfrac{1}{4}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x=5\\y^2=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=\pm\dfrac{1}{2}\end{matrix}\right.\)
Vậy ...
Chúc bạn học tốt!
bài 1) ta có : \(\dfrac{2x-y}{x+y}=\dfrac{2}{3}\Leftrightarrow2\left(x+y\right)=3\left(2x-y\right)\)
\(\Leftrightarrow2x+2y=6x-3y\Leftrightarrow4x=5y\Leftrightarrow\dfrac{x}{y}=\dfrac{5}{4}\)
vậy \(\dfrac{x}{y}=\dfrac{5}{4}\)
bài 1
\(\dfrac{2x-y}{x+y}=\dfrac{2}{3}\Leftrightarrow\dfrac{2.\dfrac{x}{y}-1}{\dfrac{x}{y}+1}=\dfrac{2.\dfrac{x}{y}+2-3}{\dfrac{x}{y}+1}=2-\dfrac{3}{\dfrac{x}{y}+1}=\dfrac{2}{3}\)
\(2-\dfrac{2}{3}=\dfrac{4}{3}=\dfrac{3}{\dfrac{x}{y}+1}\)
\(\left(\dfrac{x}{y}+1\right)=\dfrac{9}{4}\Rightarrow\dfrac{x}{y}=\dfrac{9}{4}-\dfrac{4}{4}=\dfrac{5}{4}\)
\(\left(x-\dfrac{1}{2}\right)^2+\left(y+\dfrac{1}{2}\right)^2=0\)
Nhận xét:
\(\left\{{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2\ge0,\forall x\\\left(y+\dfrac{1}{2}\right)^2\ge0,\forall y\end{matrix}\right.\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\left(y+\dfrac{1}{2}\right)^2\ge0,\forall x,y\)
Dấu \("="\) xảy ra khi:
\(\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\y+\dfrac{1}{2}=0\end{matrix}\right.\)
\(\Rightarrow x=\dfrac{1}{2};y=-\dfrac{1}{2}\)
Vậy \(x=\dfrac{1}{2};y=-\dfrac{1}{2}\)
\(\left(x-\dfrac{1}{2}\right)^2+\left(y+\dfrac{1}{2}\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2=0\\\left(y+\dfrac{1}{2}\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\)