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30 tháng 11 2019

Violympic toán 9

1 tháng 12 2019

Violympic toán 9

3 tháng 5 2019

1)

a) \(\left\{{}\begin{matrix}2x-y=5\\x+y=4\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}2x-y+x+y=5+4\\x+y=4\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}3x=9\\x+y=4\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)

Vậy (x;y)=(3;1)

b) \(16x^5-8x^3+x=0\Leftrightarrow x\left(16x^4-8x^2+1\right)=0\Leftrightarrow x\left[\left(4x^2\right)^2-2.4x^2.1+1^2\right]=0\Leftrightarrow x\left(4x^2-1\right)^2=0\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\4x^2-1=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\x=\frac{\pm1}{2}\end{matrix}\right.\)

Vậy S={\(-\frac{1}{2};0;\frac{1}{2}\)}

2)

A=\(\frac{\sqrt{\left(\sqrt{5}-1\right)^2}}{4}+\frac{1}{\sqrt{5}-1}=\frac{\sqrt{5}-1}{4}+\frac{\sqrt{5}+1}{5-1}=\frac{\sqrt{5}-1}{4}+\frac{\sqrt{5}+1}{4}=\frac{\sqrt{5}-1+\sqrt{5}+1}{4}=\frac{2\sqrt{5}}{4}=\frac{\sqrt{5}}{2}\)

B=\(\frac{4}{3+\sqrt{5}}-\frac{8}{1+\sqrt{5}}+\frac{15}{\sqrt{5}}=\frac{4\left(3-\sqrt{5}\right)}{9-5}-\frac{8\left(1-\sqrt{5}\right)}{1-5}+3\sqrt{5}=\frac{4\left(3-\sqrt{5}\right)}{4}-\frac{8\left(\sqrt{5}-1\right)}{4}+3\sqrt{5}=3-\sqrt{5}-2\sqrt{5}+2+3\sqrt{5}=5\)

30 tháng 8 2017

Áp dụng phương pháp tập thể dục

\(2-\frac{x-1}{x}=\left(\frac{\sqrt[3]{2x^2+x^3}+x+2}{2x+1}\right)^2\)

\(\Leftrightarrow\frac{x+1}{x}=\frac{\sqrt[3]{\left(2x^2+x^3\right)^2}+2\left(x+2\right)\sqrt[3]{2x^2+x^3}+\left(x+2\right)^2}{\left(2x+1\right)^2}\)

\(\Leftrightarrow\sqrt[3]{\left(2x^2+x^3\right)^2}+2\left(x+2\right)\sqrt[3]{2x^2+x^3}+\left(x+2\right)^2-\frac{\left(x+1\right)\left(2x+1\right)^2}{x}=0\)

\(\Leftrightarrow\left(\sqrt[3]{\left(2x^2+x^3\right)^2}-1\right)+2\left(x+2\right)\left(\sqrt[3]{2x^2+x^3}-1\right)+1+2\left(x+2\right)+\left(x+2\right)^2-\frac{\left(x+1\right)\left(2x+1\right)^2}{x}=0\)

\(\Leftrightarrow\frac{\left(x^2+x-1\right)\left(x^4+3x^3+2x^2+x+1\right)}{\sqrt[3]{\left(2x^2+x^3\right)^4}+\sqrt[3]{\left(2x^2+x^3\right)^2}+1}+\frac{2\left(x+2\right)\left(x+1\right)\left(x^2+x-1\right)}{\sqrt[3]{\left(2x^2+x^3\right)^2}+\sqrt[3]{2x^2+x^3}+1}+\frac{\left(1-3x\right)\left(x^2+x-1\right)}{x}=0\)

\(\Leftrightarrow\left(x^2+x-1\right)\left(\frac{\left(x^4+3x^3+2x^2+x+1\right)}{\sqrt[3]{\left(2x^2+x^3\right)^4}+\sqrt[3]{\left(2x^2+x^3\right)^2}+1}+\frac{2\left(x+2\right)\left(x+1\right)}{\sqrt[3]{\left(2x^2+x^3\right)^2}+\sqrt[3]{2x^2+x^3}+1}+\frac{\left(1-3x\right)}{x}\right)=0\)

\(\Leftrightarrow x^2+x-1=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1+\sqrt{5}}{2}\\x=\frac{-1-\sqrt{5}}{2}\end{cases}}\)

30 tháng 8 2017

@@ hoa mắt alibaba nguyễn

22 tháng 6 2016

\(\frac{\left(x+4\right)\left(x-2\right)}{x^2-2x+3}=\left(x+1\right)\frac{x+2-4}{\sqrt{x+2}+2}\)

\(\left(x-2\right)\left(\frac{x+4}{x^2-2x+3}-\frac{x+1}{\sqrt{x+2}+2}\right)=0\)

+ x=2

+ chiu kho lam cai con lai