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Câu 1:
Nhận thấy \(x=0\) không phải nghiệm, pt tương đương:
\(\frac{\left(x^2+\frac{1}{4}+3x\right)}{x}.\frac{\left(x^2+\frac{1}{4}-x\right)}{x}=12\)
\(\Leftrightarrow\left(x+\frac{1}{4x}+3\right)\left(x+\frac{1}{4x}-1\right)-12=0\)
Đặt \(x+\frac{1}{4x}-1=a\) ta được:
\(\left(a+4\right)a-12=0\Leftrightarrow a^2+4a-12=0\) \(\Rightarrow\left[{}\begin{matrix}a=2\\a=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{4x}-1=2\\x+\frac{1}{4x}-1=-6\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-3x+\frac{1}{4}=0\\x^2+5x+\frac{1}{4}=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}...\\...\end{matrix}\right.\)
Câu 2:
\(x=\sqrt{3+\sqrt{12+2\sqrt{12}+1}}=\sqrt{3+\sqrt{\left(\sqrt{12}+1\right)^2}}\)
\(=\sqrt{4+\sqrt{12}}=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
\(y=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\Rightarrow\sqrt{y}=\sqrt{3}-1\)
\(B=\frac{2\left(4+2\sqrt{3}\right)-5\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)+3\left(4-2\sqrt{3}\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)-\left(4-2\sqrt{3}\right)}\)
\(B=\frac{8+4\sqrt{3}-10+12-6\sqrt{3}}{2-4+2\sqrt{3}}=\frac{10-2\sqrt{3}}{-2+2\sqrt{3}}=\frac{5-\sqrt{3}}{\sqrt{3}-1}\)
\(B=\frac{\left(5-\sqrt{3}\right)\left(\sqrt{3}+1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\frac{5\sqrt{3}+5-3-\sqrt{3}}{2}=\frac{2+4\sqrt{3}}{2}=2\sqrt{3}+1\)
1) ĐK: \(x\ge0\)
PT \(\Leftrightarrow\frac{2}{3}\sqrt{12x}+\sqrt{12x}-\frac{1}{3}\sqrt{3x}=9\)
\(\Leftrightarrow\frac{5}{3}\sqrt{12x}-\frac{1}{3}\sqrt{3x}=9\)
\(\Leftrightarrow3\sqrt{3x}=9\) \(\Leftrightarrow x=3\left(TM\right)\)
Vậy \(x=3\)
2) ĐK: \(x\ge0\)
PT \(\Leftrightarrow7\sqrt{2x}=14\) \(\Leftrightarrow x=2\left(TM\right)\)
Vậy \(x=2\)
a) VT bạn bình phương rồi B.C.S sẽ được VT<=2
VP=3x^2-12x+12+2=3(x-2)^2+1>=2
Dấu = xảy ra khi x=2
\(\text{Đk: }1,5\le x\le2,5\)
Áp dụng bđt cauchy ta có:
\(\text{VT }\Leftrightarrow\frac{2x-3+1+1-2x+1}{2}=2\)
Mà: \(\text{VP}=3\left(x-2\right)^2+2\ge2\)
\(\text{ĐT}\Leftrightarrow x=2\)
\(\Rightarrow x=2\)
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=\(\frac{4}{3+\sqrt{5}}-\frac{8}{1+\sqrt{5}}+\frac{15}{\sqrt{5}}=\frac{4\left(3-\sqrt{5}\right)}{9-5}-\frac{8\left(1-\sqrt{5}\right)}{1-5}+3\sqrt{5}=\frac{4\left(3-\sqrt{5}\right)}{4}-\frac{8\left(\sqrt{5}-1\right)}{4}+3\sqrt{5}=3-\sqrt{5}-2\sqrt{5}+2+3\sqrt{5}=5\)
\(\sqrt{2x+1}-\sqrt{3x}=x-1\)
ĐK: \(x\ge0\)
\(\sqrt{2x+1}-\sqrt{3x}=3x-\left(2x+1\right)\)
\(\Leftrightarrow\sqrt{2x+1}-\sqrt{3x}=\left(\sqrt{3x}-\sqrt{2x+1}\right)\left(\sqrt{3x}+\sqrt{2x+1}\right)\)
\(\Leftrightarrow\left(\sqrt{2x+1}-\sqrt{3x}\right)\left(1+\sqrt{3x}+\sqrt{2x+1}\right)=0\)
\(\Leftrightarrow\sqrt{2x+1}=\sqrt{3x}\Rightarrow x=1\left(tm\right)\)
Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen
help me, pleaseee
Cần gấp lắm ạ!