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\(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
<=> x + 1 = 16
<=> x = 15 (nhận)
~ ~ ~
\(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow\sqrt{x+5}=2\)
<=> x + 5 = 4
<=> x = - 1 (nhận)
Tại x=15\(\Rightarrow\sqrt{15x^2-8x+\sqrt{15}+16}=\sqrt{15.\left(\sqrt{15}\right)^2-8.\sqrt{15}.\sqrt{15}+16}=\sqrt{15^2-2.15.4+4^2}=\sqrt{\left(15-4\right)^2}=\sqrt{11^2}=11\)
Ta có \(\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}-\dfrac{6+2\sqrt{6}}{\sqrt{3}+\sqrt{2}}=\dfrac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}+\dfrac{2\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}=\sqrt{3}+2\sqrt{3}=3\sqrt{3}\)
chắc bạn chép sai đề rồi , hai căn đầu phải 1 cộng 1 trừ chứ
a) \(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}=\sqrt{3}-1-\sqrt{3}=-1\)
b) \(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}=\sqrt{\left(3+\sqrt{2}\right)^2}-3+\sqrt{2}=3+\sqrt{2}-3+\sqrt{2}\)
\(=2\sqrt{2}\)
c) \(x-4+\sqrt{16-8x+x^2}=x-4+\sqrt{\left(x-4\right)^2}=x-4+\left|x-4\right|\)
\(=x-4+x-4\left(x>4\right)=2x-8\)
d) \(\dfrac{x^2-5}{x+\sqrt{5}}=\dfrac{\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)}{x+\sqrt{5}}=x-\sqrt{5}\)
e) \(\dfrac{x^2+2\sqrt{2}x+2}{x+\sqrt{2}}=\dfrac{\left(x+\sqrt{2}\right)^2}{x+\sqrt{2}}=x+\sqrt{2}\)
g) \(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}=\dfrac{1}{\sqrt{2}}\)
\(ĐK:x\ge0\\ PT\Leftrightarrow\dfrac{5}{3}\sqrt{15x}-\sqrt{15x}-\dfrac{1}{3}\sqrt{15x}=2\\ \Leftrightarrow\sqrt{15x}\left(\dfrac{5}{3}-1-\dfrac{1}{3}\right)=2\\ \Leftrightarrow\dfrac{1}{3}\sqrt{15x}=2\Leftrightarrow\sqrt{15x}=6\Leftrightarrow15x=36\\ \Leftrightarrow x=\dfrac{12}{5}\left(tm\right)\)
\(ĐK:x\ge0\\ PT\Leftrightarrow\dfrac{2}{3}\sqrt{15x}-\dfrac{1}{3}\sqrt{15x}=2\\ \Leftrightarrow\dfrac{1}{3}\sqrt{15x}=2\Leftrightarrow\sqrt{15x}=6\\ \Leftrightarrow15x=36\Leftrightarrow x=\dfrac{12}{5}\left(tm\right)\)
1.
\(\frac{3\sqrt{5}-5\sqrt{3}}{\sqrt{15}-3}=\frac{3\sqrt{5}-\sqrt{5}.\sqrt{15}}{\sqrt{15}-3}=\frac{-\sqrt{5}(\sqrt{15}-3)}{\sqrt{15}-3}=-\sqrt{5}\)
2.
\(\frac{\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{3}}=\frac{\sqrt{2+2\sqrt{2.3}+3}}{\sqrt{2}+\sqrt{3}}=\frac{\sqrt{(\sqrt{2}+\sqrt{3})^2}}{\sqrt{2}+\sqrt{3}}\)
\(=\frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}+\sqrt{3}}=1\)
3.
\(\frac{7+4\sqrt{3}}{2+\sqrt{3}}=\frac{2^2+2.2\sqrt{3}+3}{2+\sqrt{3}}=\frac{(2+\sqrt{3})^2}{2+\sqrt{3}}=2+\sqrt{3}\)
4.
\(\frac{16-6\sqrt{7}}{\sqrt{7}-3}=\frac{3^2-2.3\sqrt{7}+7}{\sqrt{7}-3}=\frac{(\sqrt{7}-3)^2}{\sqrt{7}-3}=\sqrt{7}-3\)
5.
\(\frac{(\sqrt{3}-\sqrt{2})^2+4\sqrt{6}}{\sqrt{3}+\sqrt{2}}=\frac{3+2+2\sqrt{2.3}}{\sqrt{3}+\sqrt{2}}=\frac{(\sqrt{3}+\sqrt{2})^2}{\sqrt{3}+\sqrt{2}}=\sqrt{3}+\sqrt{2}\)
6.
\(=\frac{(\sqrt{3})^2+(2\sqrt{5})^2-2.\sqrt{3}.2\sqrt{5}}{\sqrt{6-2\sqrt{10}}}=\frac{(\sqrt{3}-2\sqrt{5})^2}{\sqrt{6-2\sqrt{10}}}\)
\(M=\sqrt{15x^2-8x\sqrt{15+16}}\)
\(\Leftrightarrow M=\sqrt{\left(\sqrt{15}.x+4\right)^2}=\left|\sqrt{15}.x+4\right|=\sqrt{15}.x+4\)
Thay \(x=\sqrt{\dfrac{3}{5}}+\sqrt{\dfrac{5}{3}}\) vào BT ta được:
\(M=\sqrt{15}.\left(\sqrt{\dfrac{3}{5}}+\sqrt{\dfrac{5}{3}}\right)+4=\sqrt{9}+\sqrt{25}+4=3+5+4=12\)
Vậy \(M=12\)
thanks bạn nha :))