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\(\frac{x}{-15}=\frac{-60}{x}\Leftrightarrow\frac{x}{-1}=\frac{-4}{x}\Leftrightarrow x\cdot x=4\)
\(\Leftrightarrow x^2=4\Leftrightarrow x=\pm2\)
\(\frac{-2}{x}=\frac{-x}{\frac{8}{25}}\)
\(\Leftrightarrow x\cdot(-x)=-2\cdot\frac{8}{25}\)
\(\Leftrightarrow-x^2=-\frac{16}{25}\)
\(\Leftrightarrow x^2=\frac{16}{25}\Leftrightarrow x=\pm\frac{4}{5}\)
\(\frac{x}{-15}=\frac{-60}{x}\)
\(\Rightarrow x^2=900\)
\(\Rightarrow x=\pm30\)
\(\frac{-2}{x}=\frac{-x}{\frac{8}{25}}\)
\(\Rightarrow-\left(x^2\right)=-\frac{16}{25}\)
\(\Rightarrow-\left(x^2\right)=-\left(\frac{4}{5}\right)^2\)
\(\Rightarrow x=\frac{4}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,\(\frac{x}{-15}=\frac{-60}{x}\)=\(x^2\)= [-15 ] . [60] = 900 \(\Rightarrow\)x = + 30
a.
\(\frac{x}{-15}=\frac{-60}{x}=x^2=\left(-15\right):\left(60\right)=900\Rightarrow x=+30\)
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1.a) Sửa lại đề: \(\frac{11}{17}\)ở mẫu chuyển thành \(\frac{11}{7}\)
\(\frac{0,75+0,6-\frac{3}{7}-\frac{3}{13}}{2,75+2,2-\frac{11}{7}-\frac{11}{13}}=\frac{\frac{3}{4}+\frac{3}{5}-\frac{3}{7}-\frac{3}{13}}{\frac{11}{4}+\frac{11}{5}-\frac{11}{7}-\frac{11}{13}}\)\(=\frac{3\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\right)}{11\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\right)}=\frac{3}{11}\)
( vì \(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\ne0\))
2.a) \(\frac{3}{5}+\frac{3}{2}.x=\frac{-5}{7}\)\(\Leftrightarrow\frac{3}{2}.x=\frac{-5}{7}-\frac{3}{5}\)
\(\Leftrightarrow\frac{3}{2}.x=\frac{-46}{35}\)\(\Leftrightarrow x=\frac{-46}{35}:\frac{3}{2}\)\(\Leftrightarrow x=\frac{-92}{105}\)
Vậy \(x=\frac{-92}{105}\)
b) \(\left(4x-\frac{1}{3}\right).\left(\frac{3}{2}x+\frac{5}{6}\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}4x-\frac{1}{3}=0\\\frac{3}{2}x+\frac{5}{6}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}4x=\frac{1}{3}\\\frac{3}{2}x=\frac{-5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{12}\\x=\frac{-5}{9}\end{cases}}\)
Vậy \(x=\frac{-5}{9}\)hoặc \(x=\frac{1}{12}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a.
\(\frac{x}{-15}=-\frac{60}{x}\)
\(x\times x=15\times60\)
\(x^2=900\)
\(x^2=\left(\pm30\right)^2\)
\(x=\pm30\)
Vậy \(x=-30\) hoặc \(x=30\)
b.
\(-\frac{2}{x}=\frac{-x}{\frac{8}{25}}\)
\(x\times x=2\times\frac{8}{25}\)
\(x^2=\frac{16}{25}\)
\(x^2=\left(\pm\frac{4}{5}\right)^2\)
\(x=\pm\frac{4}{5}\)
Vậy \(x=\frac{4}{5}\) hoặc \(x=-\frac{4}{5}\)
Chúc bạn học tốt ^^
a/Theo bài ra ta có x.x= (-15).(-60)= \(x^2\)=900 mà 900= \(30^2\)suy ra x=30
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a) \(\frac{x-2}{5}=\frac{3}{8}\)
(x-2).8=5.3
(x-2).8=15
x-2=15:8
x-2=\(\frac{15}{8}\)
x=\(\frac{15}{8}+2\)
x=\(\frac{31}{8}\)
b)\(\frac{x-1}{x+5}=\frac{6}{7}\)
(x-1).7=(x+5).6
7x-7=6x+30
7x=6x+30+7
7x=6x+37
7x-6x=37
x=37
c)\(\frac{x^2}{6}=\frac{24}{25}\)
\(x^2.25=6.24\)
\(x^2.25=144\)
\(x^2=144:25\)
\(x^2=\frac{144}{25}\)
\(x^2=\left(\frac{12}{5}\right)^2\)
\(x=\frac{12}{5}\)
=\(2\times\frac{-8}{25}=x^2\)
\(\rightarrow\frac{-16}{25}=x^2\)
ĐỀ BÀI CỐ VẤN ĐỀ J KO BẠN ƠI
- Xét x < 0 thì \(\frac{2}{x}< \frac{-x}{\frac{8}{25}}\)\(\Rightarrow\)x không thoả mãn
- Xét x = 0 thì\(\frac{2}{x}\)không có giá trị ( vì không có số nào chia được cho 0 ) vậy x không thoả mãn
- Xét x > 0 thì \(\frac{2}{x}>\frac{-x}{\frac{8}{25}}\)\(\Rightarrow\)x không thoả mãn
\(\Rightarrow x\in\varnothing\)