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Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen
help me, pleaseee
Cần gấp lắm ạ!
A = \(x^2+3x-7=x^2+2x\frac{3}{2}+\frac{9}{4}-\frac{37}{4}\)
\(=\left(x+\frac{3}{2}\right)^2-\frac{37}{4}\ge-\frac{37}{4}\)
\(\Rightarrow\)min A = \(-\frac{37}{4}\Leftrightarrow x=-\frac{3}{2}\)
B = \(x-5\sqrt{x}-1\) ĐKXĐ: \(x\ge0\)
\(=x-2\sqrt{x}\frac{5}{2}+\frac{25}{4}-\frac{29}{4}=\left(\sqrt{x}-\frac{5}{2}\right)^2-\frac{29}{4}\ge-\frac{29}{4}\)
\(\Rightarrow\)min B = \(-\frac{29}{4}\Leftrightarrow x=\frac{25}{4}\)( thỏa mãn)
C = \(\frac{-4}{\sqrt{x}+7}\) ĐKXĐ:\(x\ge0\)
Ta có: \(\sqrt{x}+7\ge7\Rightarrow\frac{4}{\sqrt{x}+7}\le\frac{4}{7}\)\(\Leftrightarrow\frac{-4}{\sqrt{x}+7}\ge-\frac{4}{7}\)
\(\Rightarrow\)min C = \(-\frac{4}{7}\Leftrightarrow x=0\)
D = \(\frac{\sqrt{x}+1}{\sqrt{x}+3}\) ĐKXĐ:\(x\ge0\)
\(=1-\frac{2}{\sqrt{x}+3}\ge1-\frac{2}{3}=\frac{1}{3}\)
\(\Rightarrow\)min D = \(\frac{1}{3}\Leftrightarrow x=0\)
E = \(\frac{x+7}{\sqrt{x}+3}\) ĐKXĐ:\(x\ge0\)
\(=\frac{x-9+16}{\sqrt{x}+3}=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)+16}{\sqrt{x}+3}=\sqrt{x}-3+\frac{16}{\sqrt{x}+3}=\sqrt{x}+3+\frac{16}{\sqrt{x}+3}-6\ge2\sqrt{16}-6=2\)
\(\Rightarrow\)min E = \(2\Leftrightarrow x=1\)(thỏa mãn)
F = \(\frac{x^2+3x+5}{x^2}\) ĐKXĐ: \(x\ne0\)
\(\Leftrightarrow\)\(x^2\left(F-1\right)-3x-5=0\)
△ = \(3^2+20\left(F-1\right)\ge0\)\(\Leftrightarrow F\ge\frac{11}{20}\)
\(\Rightarrow\)min F = \(\frac{11}{20}\Leftrightarrow x=-\frac{10}{3}\)( thỏa mãn)
Bài 1:
a)
\(\sqrt{13-2\sqrt{42}}=\sqrt{6+7-2\sqrt{6.7}}=\sqrt{(\sqrt{7}-\sqrt{6})^2}=|\sqrt{7}-\sqrt{6}|=\sqrt{7}-\sqrt{6}\)
b)
\(\sqrt{46+6\sqrt{5}}=\sqrt{46+2\sqrt{45}}=\sqrt{45+1+2\sqrt{45.1}}=\sqrt{(\sqrt{45}+1)^2}=\sqrt{45}+1\)
\(=3\sqrt{5}+1\)
c)
\(\sqrt{12-3\sqrt{15}}=\sqrt{\frac{24-6\sqrt{15}}{2}}=\sqrt{\frac{24-2\sqrt{135}}{2}}=\sqrt{\frac{15+9-2\sqrt{15.9}}{2}}\)
\(=\sqrt{\frac{(\sqrt{15}-\sqrt{9})^2}{2}}=\frac{\sqrt{15}-\sqrt{9}}{\sqrt{2}}=\frac{\sqrt{15}-3}{\sqrt{2}}\)
d)
\(\sqrt{11+\sqrt{96}}=\sqrt{11+2\sqrt{24}}=\sqrt{8+3+2\sqrt{8.3}}\)
\(=\sqrt{(\sqrt{8}+\sqrt{3})^2}=\sqrt{8}+\sqrt{3}\)
Bài 1:
a)
\(\sqrt{13-2\sqrt{42}}=\sqrt{6+7-2\sqrt{6.7}}=\sqrt{(\sqrt{7}-\sqrt{6})^2}=|\sqrt{7}-\sqrt{6}|=\sqrt{7}-\sqrt{6}\)
b)
\(\sqrt{46+6\sqrt{5}}=\sqrt{46+2\sqrt{45}}=\sqrt{45+1+2\sqrt{45.1}}=\sqrt{(\sqrt{45}+1)^2}=\sqrt{45}+1\)
\(=3\sqrt{5}+1\)
c)
\(\sqrt{12-3\sqrt{15}}=\sqrt{\frac{24-6\sqrt{15}}{2}}=\sqrt{\frac{24-2\sqrt{135}}{2}}=\sqrt{\frac{15+9-2\sqrt{15.9}}{2}}\)
\(=\sqrt{\frac{(\sqrt{15}-\sqrt{9})^2}{2}}=\frac{\sqrt{15}-\sqrt{9}}{\sqrt{2}}=\frac{\sqrt{15}-3}{\sqrt{2}}\)
d)
\(\sqrt{11+\sqrt{96}}=\sqrt{11+2\sqrt{24}}=\sqrt{8+3+2\sqrt{8.3}}\)
\(=\sqrt{(\sqrt{8}+\sqrt{3})^2}=\sqrt{8}+\sqrt{3}\)
Bài 4 :
\(a,\sqrt{x-1}=2\)
=> \(x-1=2^2=4\)
=>\(x=4+1=5\)
Vậy \(x\in\left\{5\right\}\)
\(b,\sqrt{x^2-3x+2}=2\)
=> \(x^2-3x+2=2\)
=> \(x^2-3x=2-2=0\)
=>\(x.\left(x-3\right)=0\)( phân tích đa thức thanh nhân tử )
=> \(\left[{}\begin{matrix}x=0\\x-3=0=>x=0+3=3\end{matrix}\right.\)
Vậy \(x\in\left\{0;3\right\}\)
MÌNH Biết vậy thôi ,
Bài 4 :
c) \(\sqrt{4x+1}=x+1\)ĐK : \(x\ge-1\)
\(\Leftrightarrow4x+1=\left(x+1\right)^2\)
\(\Leftrightarrow x^2+2x+1-4x-1=0\)
\(\Leftrightarrow x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)( thỏa )
d) \(\sqrt{x+2\sqrt{x-1}}-\sqrt{x-2\sqrt{x-1}}=2\)
\(\Leftrightarrow\sqrt{x-1+2\sqrt{x-1}+1}-\sqrt{x-1-2\sqrt{x-1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}-\sqrt{\left(\sqrt{x-1}-1\right)^2}=2\)
\(\Leftrightarrow\left|\sqrt{x-1}+1\right|-\left|\sqrt{x-1}-1\right|=2\)
+) Xét \(x\ge2\)
\(pt\Leftrightarrow\sqrt{x-1}+1-\sqrt{x-1}+1=2\)
\(\Leftrightarrow2=2\)( luôn đúng )
+) Xét \(1\le x< 2\):
\(pt\Leftrightarrow\sqrt{x-1}+1-1+\sqrt{x-1}=2\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\)( loại )
Vậy \(x\ge2\)