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\(\frac{x-2}{4}=\frac{-16}{2-x}\)
\(\Leftrightarrow\frac{x-2}{4}=\frac{16}{x-2}\)
\(\Leftrightarrow\left(x-2\right)^2=64\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=8\\x-2=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}x=10\\x=-6\end{cases}}}\)
Vậy x=10; x=-6
Ta có:
\(\frac{x}{2}=\frac{y}{3}=\frac{x}{8}=\frac{y}{12}\left(1\right)\)
\(\frac{y}{4}=\frac{z}{5}=\frac{y}{12}=\frac{z}{15}\left(2\right)\)
từ (1) và (2) => \(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x^2}{64}=\frac{y^2}{144}=\frac{z}{15}\)
\(=\frac{x^2-y^2}{64-144}=\frac{-16}{-80}=\frac{1}{5}\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{8}=\frac{1}{5}\Rightarrow x=\frac{8}{5}\\\frac{y}{12}=\frac{1}{5}\Rightarrow y=\frac{12}{5}\\\frac{z}{15}=\frac{1}{5}\Rightarrow z=3\end{cases}}\)
Vậy \(\hept{\begin{cases}x=\frac{8}{5}\\y=\frac{12}{5}\\z=3\end{cases}}\)
ta có:
\(\frac{x}{2}=\frac{y}{3}=\frac{x}{8}=\frac{y}{12}\)(1)
\(\frac{y}{4}=\frac{z}{5}=\frac{y}{12}=\frac{z}{15}\)(2)
từ (1) và (2) ta được :\(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}\)
áp dụng tính chất của dãy tỷ số bằng nhau ta có:
\(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x}{8^2}^2=\frac{y^2}{12^2}=\frac{x^2-y^2}{64-144}=-\frac{16}{80}=\frac{1}{5}\)
\(\Rightarrow\frac{x}{8}=\frac{1}{5}\Rightarrow x=\frac{8}{5}\)
\(\Rightarrow\frac{y}{12}=\frac{1}{5}=>y=\frac{12}{5}\)
\(=>\frac{z}{15}=\frac{1}{5}=>z=3\)
vậy/.....
chúc bn học tốt !!
Ta có: \(\hept{\begin{cases}\frac{x}{2}=\frac{y}{3};\frac{y}{4}=\frac{z}{5}\\x^2-y^2=-16\end{cases}\Rightarrow\frac{x}{8}=\frac{y}{12}=\frac{z}{15}}\)
\(\Rightarrow\frac{x^2}{64}=\frac{y^2}{144}=\frac{z^2}{225}=\frac{x^2-y^2}{64-144}=\frac{-16}{-80}=\frac{1}{5}\)
\(\Rightarrow x^2=\frac{1}{5}.64=\frac{64}{5}\Rightarrow x=+_-\sqrt{\frac{64}{5}}\)
\(y^2=\frac{1}{5}.144=\frac{144}{5}\Rightarrow y=+_-\sqrt{\frac{144}{5}}\)
\(z^2=\frac{1}{5}.255=51\Rightarrow z=+_-\sqrt{51}\)
CHÚC BẠN HỌC TỐT
a ) \(\left(\frac{2}{5}-x\right):1\frac{1}{3}+\frac{1}{2}=-4\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}+\frac{1}{2}=-4\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}=-4-\frac{1}{2}\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}=-\frac{9}{2}\)
\(\frac{2}{5}-x=-\frac{9}{2}.\frac{4}{3}\)
\(\frac{2}{5}-x=-3\)
\(x=\frac{2}{5}-\left(-3\right)\)
\(x=\frac{2}{5}+3\)
\(x=\frac{3}{5}-\frac{15}{5}\)
\(x=-\frac{12}{5}\)
Vay \(x=-\frac{12}{5}\)
b ) \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(1+\frac{2}{5}+\frac{2}{3}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15}{15}+\frac{6}{15}+\frac{10}{15}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15+6+10}{15}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\frac{31}{15}=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{5}{4}.\frac{31}{15}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{1}{4}.\frac{31}{3}\)
\(-3+\frac{3}{x}-\frac{1}{3}=-\frac{31}{12}\)
\(-3+\frac{3}{x}=-\frac{31}{12}+\frac{1}{2}\)
\(-3+\frac{3}{x}=-\frac{31}{12}+\frac{6}{12}\)
\(-3+\frac{3}{x}=\frac{-25}{12}\)
\(\frac{3}{x}=\frac{-25}{12}+3\)
\(\frac{3}{x}=\frac{-25}{12}+\frac{36}{12}\)
\(\frac{3}{x}=\frac{5}{6}\)
\(\frac{18}{6x}=\frac{5x}{6x}\)
Đèn dây , bạn tự làm tiếp nhé , de rồi chứ
Câu b) tạm thời ko bít làm =.=
Bài 1 :
\(d)\) \(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=2x\)
\(\Leftrightarrow\)\(\frac{4^5.4}{3^5.3}.\frac{6^5.6}{2^5.2}=2x\)
\(\Leftrightarrow\)\(\frac{4^6}{3^6}.\frac{6^6}{2^6}=2x\)
\(\Leftrightarrow\)\(\frac{2^{12}}{3^6}.\frac{2^6.3^6}{2^6}=2x\)
\(\Leftrightarrow\)\(\frac{2^{12}}{3^6}.\frac{3^6}{1}=2x\)
\(\Leftrightarrow\)\(2^{12}=2x\)
\(\Leftrightarrow\)\(x=\frac{2^{12}}{2}\)
\(\Leftrightarrow\)\(x=2^{11}\)
\(\Leftrightarrow\)\(x=2048\)
Vậy \(x=2048\)
Chúc bạn học tốt ~
Bài 1 :
\(a)\) Ta có :
\(4+\frac{x}{7+y}=\frac{4}{7}\)
\(\Leftrightarrow\)\(\frac{x}{7+y}=\frac{4}{7}-4\)
\(\Leftrightarrow\)\(\frac{x}{7+y}=\frac{-24}{7}\)
\(\Leftrightarrow\)\(\frac{x}{-24}=\frac{7+y}{7}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{-24}=\frac{7+y}{7}=\frac{x+7+y}{-24+7}=\frac{22+7}{-17}=\frac{29}{-17}=\frac{-29}{17}\)
Do đó :
\(\frac{x}{-24}=\frac{-29}{17}\)\(\Rightarrow\)\(x=\frac{-29}{17}.\left(-24\right)=\frac{696}{17}\)
\(\frac{7+y}{7}=\frac{-29}{17}\)\(\Rightarrow\)\(y=\frac{-29}{17}.7-7=\frac{-322}{17}\)
Vậy \(x=\frac{696}{17}\) và \(y=\frac{-322}{17}\)
Chúc bạn học tốt ~
1.b) \(\left(\left|x\right|-3\right)\left(x^2+4\right)< 0\)
\(\Rightarrow\hept{\begin{cases}\left|x\right|-3\\x^2+4\end{cases}}\) trái dấu
\(TH1:\hept{\begin{cases}\left|x\right|-3< 0\\x^2+4>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|< 3\\x^2>-4\end{cases}}\Leftrightarrow x\in\left\{0;\pm1;\pm2\right\}\)
\(TH1:\hept{\begin{cases}\left|x\right|-3>0\\x^2+4< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|>3\\x^2< -4\end{cases}}\Leftrightarrow x\in\left\{\varnothing\right\}\)
Vậy \(x\in\left\{0;\pm1;\pm2\right\}\)
Ta có :
\(\frac{x-2}{4}=\frac{-16}{2-x}\)
\(\Leftrightarrow\)\(\frac{x-2}{4}=\frac{16}{x-2}\)
\(\Leftrightarrow\)\(\left(x-2\right)\left(x-2\right)=16.4\)
\(\Leftrightarrow\)\(\left(x-2\right)^2=64\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}\left(x-2\right)^2=8^2\\\left(x-2\right)^2=\left(-8\right)^2\end{cases}\Leftrightarrow\orbr{\begin{cases}x-2=8\\x-2=-8\end{cases}}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=8+2\\x=-8+2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=10\\x=-6\end{cases}}}\)
Vậy \(x=-6\) và \(x=10\)
Chúc bạn học tốt ~
Ta có : \(\frac{x-2}{4}=\frac{-16}{2-x}\)
\(\Rightarrow\left(x-2\right).\left(2-x\right)=-16.4\)
\(\Rightarrow\left(x-2\right).2-\left(x-2\right).x=-64\)
\(\Rightarrow2x-4-x^2-2x=-64\)
\(\Rightarrow-4-x^2=-64\)
\(\Rightarrow x^2=-4+64\)
\(\Rightarrow x^2=60\)