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3.
ĐKXĐ: \(x\ge-1;x\ne13\)
\(\left(x+2\right)\left(\sqrt{x+1}-2\right)=\sqrt[3]{2x+1}-3\)
\(\Leftrightarrow\left(x+2\right)\sqrt{x+1}-2x-4=\sqrt[3]{2x+1}-3\)
\(\Leftrightarrow\left(x+1\right)\sqrt{x+1}+x+1-\left(2x+1\right)-\sqrt[3]{2x+1}=0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\\\sqrt[3]{2x+1}=b\end{matrix}\right.\)
\(\Rightarrow a^3+a-b^3-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+1\right)=0\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow\sqrt{x+1}=\sqrt[3]{2x+1}\) (\(x\ge-\frac{1}{2}\))
\(\Leftrightarrow\left(x+1\right)^3=\left(2x+1\right)^2\)
\(\Leftrightarrow x=?\)
2.
ĐKXĐ: \(x\ge-\frac{1}{2}\)
\(\Leftrightarrow8x^3+2x-\left(2x+2\right)\sqrt{2x+1}=0\)
Đặt \(\left\{{}\begin{matrix}2x=a\\\sqrt{2x+1}=b\end{matrix}\right.\)
\(\Rightarrow a^3+a-\left(b^2+1\right)b=0\)
\(\Leftrightarrow a^3-b^3+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+1\right)=0\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow2x=\sqrt{2x+1}\) (\(x\ge0\))
\(\Leftrightarrow4x^2=2x+1\)
\(\Leftrightarrow x=?\)
Ai phát hiện sai đề thì sửa và làm giúp mk hộ với, cảm ơn :) (chỉ cần làm tóm tắt thôi)
Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen
help me, pleaseee
Cần gấp lắm ạ!
1) Ta có: \(\left(\sqrt{12}-6\sqrt{3}+\sqrt{24}\right)\cdot\sqrt{6}-\left(\frac{5}{2}\sqrt{2}+12\right)\)
\(=\left(2\sqrt{3}-6\sqrt{3}+2\sqrt{6}\right)\cdot\sqrt{6}-\left(\sqrt{\frac{25}{4}\cdot2}+12\right)\)
\(=\left(-4\sqrt{3}+2\sqrt{6}\right)\cdot\sqrt{6}-\left(\sqrt{\frac{50}{4}}+12\right)\)
\(=-12\sqrt{2}+12-\frac{5\sqrt{2}}{2}-12\)
\(=\frac{-24\sqrt{2}-5\sqrt{2}}{2}\)
\(=\frac{-29\sqrt{2}}{2}\)
2) Ta có: \(\frac{26}{2\sqrt{3}+5}-\frac{4}{\sqrt{3}-2}\)
\(=\frac{26\left(5-2\sqrt{3}\right)}{\left(5+2\sqrt{3}\right)\left(5-2\sqrt{3}\right)}+\frac{4}{2-\sqrt{3}}\)
\(=\frac{26\left(5-2\sqrt{3}\right)}{25-12}+\frac{4\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=2\left(5-2\sqrt{3}\right)+4\left(2+\sqrt{3}\right)\)
\(=10-4\sqrt{3}+8+4\sqrt{3}\)
\(=18\)
3) ĐK để phương trình có nghiệm là: x≥0
Ta có: \(\sqrt{x^2-6x+9}=2x\)
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=2x\)
\(\Leftrightarrow\left|x-3\right|=2x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x\\x-3=-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-3-2x=0\\x-3+2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x-3=0\\3x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=3\\3x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)
Vậy: S={1}
4) ĐK để phương trình có nghiệm là: \(x\ge\frac{1}{2}\)
Ta có: \(\sqrt{4x^2+1}=2x-1\)
\(\Leftrightarrow\left(\sqrt{4x^2+1}\right)^2=\left(2x-1\right)^2\)
\(\Leftrightarrow4x^2+1=4x^2-4x+1\)
\(\Leftrightarrow4x^2+1-4x^2+4x-1=0\)
\(\Leftrightarrow4x=0\)
hay x=0(loại)
Vậy: S=∅
\(a,\frac{\sqrt{7x^2y^4}}{\sqrt{28x^4y^4}}\)
\(\frac{\sqrt{7}xy^2}{2\sqrt{7}x^2y^2}=\frac{1}{2x}\)
\(b,\sqrt{2x+2\sqrt{2x-1}}+\sqrt{2x+2\sqrt{2x-1}}\)
\(\sqrt{2x-1+2\sqrt{2x-1}+1}+\sqrt{2x-1+2\sqrt{2x-1}+1}\)
\(\sqrt{\left(\sqrt{2x-1}+1\right)^2}+\sqrt{\left(\sqrt{2x-1}+1\right)^2}\)
\(\left|\sqrt{2x-1}+1\right|+\left|\sqrt{2x-1}+1\right|\)
\(\sqrt{2x-1}+1+\sqrt{2x-1}+1\)
\(2\sqrt{2x-1}+2\)
\(c,\frac{1}{3}\sqrt{9x-27}+\sqrt{2x-6}-\sqrt{4x-12}=2-\sqrt{2}\)
\(\sqrt{x-3}+\sqrt{2}\sqrt{x-3}-2\sqrt{x-3}=2-\sqrt{2}\)
\(\sqrt{x-3}\left(1+\sqrt{2}-2\right)=2-\sqrt{2}\)
\(\sqrt{x-3}\left(\sqrt{2}-1\right)=\sqrt{2}\left(\sqrt{2}-1\right)\)
\(\sqrt{x-3}=\sqrt{2}\)
\(x-3=2< =>x=5\)
a) \(\frac{\sqrt{7\left(-x^2\right)y^4}}{\sqrt{28x^4y^4}}=\frac{\sqrt{7}xy^2}{2\sqrt{7}x^2y^2}=\frac{1}{2x}\)(vì x > 0)
b) \(\sqrt{2x+2\sqrt{2x-1}}+\sqrt{2x+2\sqrt{2x-1}}\)
\(=\sqrt{2x-1+2\sqrt{2x-1}+1}+\sqrt{2x-1+2\sqrt{2x-1}}\)
\(=\sqrt{\left(\sqrt{2x-1}+1\right)^2}+\sqrt{\left(\sqrt{2x+1}+1\right)^2}=\sqrt{2x-1}+1+\sqrt{2x+1}+1\)
\(=2\sqrt{2x-1}+2\)
c) ĐK: x \(\ge\)3
Ta có:: \(\frac{1}{3}\sqrt{9x-27}+\sqrt{2x-6}-\sqrt{4x-12}=2-\sqrt{2}\)
<=> \(\sqrt{x-3}+\sqrt{2}.\sqrt{x-3}-2\sqrt{x-3}=2-\sqrt{2}\)
<=> \(\sqrt{x-3}.\left(\sqrt{2}-1\right)=\sqrt{2}\left(\sqrt{2}-1\right)\)
<=> \(\sqrt{x-3}=\sqrt{2}\) <=> x - 3 = 2 <=> x = 5 (tm)