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Đặt \(\frac{x}{4}=\frac{y}{7}\) = k => x = 4k; y = 7k ( k khác 0)
Thay vào C ta được: \(C=\frac{\left(1+\sqrt{3}\right)\left(4k\right)^2.7k-\left(2-\sqrt{5}\right).4k.\left(7k\right)^2}{\left(4k\right)^3+\left(7k\right)^3}=\frac{\left(112.\left(1+\sqrt{3}\right)-196.\left(2-\sqrt{5}\right)\right).k^3}{407k^3}\)
\(C=\frac{112+112\sqrt{3}-392+196\sqrt{5}}{407}=\frac{112\sqrt{3} +196\sqrt{5}-280}{407}\)
Câu 3: đề là \(\sqrt{x+5}-\sqrt{x-2}\) hay \(\sqrt{x+5}-\sqrt{x+2}\)?
Câu 4:
ĐKXĐ: \(x\le9\)
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x-4}=a\\\sqrt{9-x}=b\end{matrix}\right.\) ta có hệ:
\(\left\{{}\begin{matrix}a-b=-1\\a^3+b^2=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}b=a+1\\a^3+b^2=5\end{matrix}\right.\)
\(\Rightarrow a^3+\left(a+1\right)^2=5\)
\(\Leftrightarrow a^3+a^2+2a-4=0\) \(\Rightarrow a=1\)
\(\Rightarrow\sqrt[3]{x-4}=1\Rightarrow x-4=1\Rightarrow x=5\)
5.
ĐKXĐ: \(x\ge-\frac{17}{16}\)
\(\Leftrightarrow8x^2-15x-23-\left(x+1\right)\sqrt{16x+17}=0\)
\(\Leftrightarrow\left(x+1\right)\left(8x-23\right)-\left(x+1\right)\sqrt{16x+17}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\8x-23=\sqrt{16x+17}\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow16x+17-2\sqrt{16x+17}-63=0\)
Đặt \(\sqrt{16x+17}=t\ge0\)
\(\Rightarrow t^2-2t-63=0\Rightarrow\left[{}\begin{matrix}t=9\\t=-7\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{16x+17}=9\Leftrightarrow x=\frac{32}{3}\)
Lời giải:
a)
\(\frac{4}{\sqrt{10}}(\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}})=\frac{4}{\sqrt{20}}(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}})\)
\(=\frac{4}{2\sqrt{5}}(\sqrt{5+1+2\sqrt{5}}+\sqrt{5+1-2\sqrt{5}})=\frac{2}{\sqrt{5}}[\sqrt{(\sqrt{5}+1)^2}+\sqrt{(\sqrt{5}-1)^2}]\)
\(=\frac{2}{\sqrt{5}}(\sqrt{5}+1+\sqrt{5}-1)=\frac{2}{\sqrt{5}}.2\sqrt{5}=4\)
b)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{8-2\sqrt{15}}=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{3+5-2\sqrt{3.5}}\)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})(\sqrt{5}-\sqrt{3})\)
\(=(4+\sqrt{15})(8-2\sqrt{15})=2(4+\sqrt{15})(4-\sqrt{15})=2(16-15)=2\)
c)
\(=\sqrt{4\sqrt{2}(\sqrt{3}+1)+8\sqrt{3}+18}=\sqrt{4\sqrt{2}(\sqrt{3}+1)+4(3+1+2\sqrt{3})+2}\)
\(=\sqrt{4\sqrt{2}(\sqrt{3}+1)+4(\sqrt{3}+1)^2+2}\)
\(=\sqrt{(2\sqrt{3}+2)^2+(\sqrt{2})^2+2.(2\sqrt{3}+2).\sqrt{2}}\)
\(=\sqrt{(2\sqrt{3}+2+\sqrt{2})^2}=2\sqrt{3}+2+\sqrt{2}\)
1)
a) Ta có : \(\frac{x^2+5}{\sqrt{x^2+4}}=\frac{\left(x^2+4\right)+1}{\sqrt{x^2+4}}=\sqrt{x^2+4}+\frac{1}{\sqrt{x^2+4}}\). Đến đây áp dụng bđt \(a+\frac{1}{a}>2\)là ra nhé :)
b) Ta sẽ chứng minh bằng biến đổi tương đương :
\(\sqrt{\left(a+c\right)\left(b+d\right)}\ge\sqrt{ab}+\sqrt{cd}\)
\(\Leftrightarrow\left(a+c\right)\left(b+d\right)\ge\left(\sqrt{ab}+\sqrt{cd}\right)^2\)
\(\Leftrightarrow ab+ad+bc+cd\ge ab+cd+2\sqrt{abcd}\)
\(\Leftrightarrow ad-2\sqrt{abcd}+bc\ge0\)
\(\Leftrightarrow\left(\sqrt{ad}-\sqrt{bc}\right)^2\ge0\)(luôn đúng)
Vì bđt cuối luôn đúng nên bđt ban đầu được chứng minh.
2) Mình làm tóm tắt thôi nhé , do đề dài...
a) \(\sqrt{2x+\sqrt{4x-1}}-\sqrt{2x-\sqrt{4x-1}}\)
\(=\frac{\sqrt{\left(4x-1\right)+2\sqrt{4x-1}+1}+\sqrt{\left(4x-1\right)-2\sqrt{4x-1}+1}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{4x-1}+1\right)^2}+\sqrt{\left(\sqrt{4x-1}+1\right)^2}}{\sqrt{2}}=\frac{\left|\sqrt{4x-1}-1\right|+\left|\sqrt{4x-1}+1\right|}{\sqrt{2}}\)
b) \(\frac{x-y+3\sqrt{x}+3\sqrt{y}}{\sqrt{x}-\sqrt{y}+3}=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)+3\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}+3}\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}+3\right)}{\sqrt{x}-\sqrt{y}+3}=\sqrt{x}+\sqrt{y}\)
c) Biến đổi : \(\sqrt{x-2\sqrt{x-1}}=\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}=\sqrt{\left(\sqrt{x-1}-1\right)^2}=\left|\sqrt{x-1}-1\right|\)
d) Biến đổi tương tự c)
e) \(\sqrt{x+\sqrt{x^2-4}}.\sqrt{x-\sqrt{x^2-4}}=\sqrt{x^2-\left(x^2-4\right)}=\sqrt{4}=2\)
a, \(=7\sqrt{2}-6\sqrt{2}+\frac{1}{2}.2\sqrt{2}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)
b, \(=4\sqrt{a}+4\sqrt{10a}-9\sqrt{10a}=4\sqrt{a}-5\sqrt{10a}\)
c, \(=6+\sqrt{15}-\sqrt{60}=6+\sqrt{15}-2\sqrt{15}=6-\sqrt{15}\)
Rút gọn
a) Ta có: \(\sqrt{98}-\sqrt{72}+\frac{1}{2}\sqrt{8}\)
\(=\sqrt{2}\left(\sqrt{49}-\sqrt{36}+\frac{1}{2}\sqrt{4}\right)\)
\(=\sqrt{2}\left(7-6+\frac{1}{2}\cdot2\right)\)
\(=\sqrt{2}\left(1+1\right)=2\sqrt{2}\)
b) Ta có: \(\sqrt{16a}+2\sqrt{40a}-3\sqrt{90a}\)
\(=\sqrt{a}\left(\sqrt{16}+2\sqrt{40}-3\sqrt{90}\right)\)
\(=\sqrt{a}\left(4+4\sqrt{10}-9\sqrt{10}\right)\)
\(=\sqrt{a}\left(4-5\sqrt{10}\right)\)
\(=4\sqrt{a}-5\sqrt{10a}\)
c) Ta có: \(\left(2\sqrt{3}+\sqrt{5}\right)\cdot\sqrt{3}-\sqrt{60}\)
\(=6+\sqrt{15}-\sqrt{60}\)
\(=6-\sqrt{15}\)
Bài 1:
\(A=\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}=\sqrt{2+3-2\sqrt{2.3}}+\sqrt{2+3+2\sqrt{2.3}}\)
\(=\sqrt{(\sqrt{2}-\sqrt{3})^2}+\sqrt{\sqrt{2}+\sqrt{3})^2}\)
\(=|\sqrt{2}-\sqrt{3}|+|\sqrt{2}+\sqrt{3}|=\sqrt{3}-\sqrt{2}+\sqrt{2}+\sqrt{3}=2\sqrt{3}\)
\(B=(\sqrt{10}+\sqrt{6})\sqrt{8-2\sqrt{15}}\)
\(=(\sqrt{10}+\sqrt{6}).\sqrt{3+5-2\sqrt{3.5}}\)
\(=(\sqrt{10}+\sqrt{6})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)
\(=\sqrt{2}(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})=\sqrt{2}(5-3)=2\sqrt{2}\)
\(C=\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}\)
\(C^2=8+2\sqrt{(4+\sqrt{7})(4-\sqrt{7})}=8+2\sqrt{4^2-7}=8+2.3=14\)
\(\Rightarrow C=\sqrt{14}\)
\(D=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{2}\sqrt{3-\sqrt{5}}\)
\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{6-2\sqrt{5}}\)
\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{5+1-2\sqrt{5.1}}\)
\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{(\sqrt{5}-1)^2}\)
\(=(3+\sqrt{5})(\sqrt{5}-1)^2=(3+\sqrt{5})(6-2\sqrt{5})=2(3+\sqrt{5})(3-\sqrt{5})=2(3^2-5)=8\)
Bài 2:
a) Bạn xem lại đề.
b) \(x-2\sqrt{xy}+y=(\sqrt{x})^2-2\sqrt{x}.\sqrt{y}+(\sqrt{y})^2=(\sqrt{x}-\sqrt{y})^2\)
c)
\(\sqrt{xy}+2\sqrt{x}-3\sqrt{y}-6=(\sqrt{x}.\sqrt{y}+2\sqrt{x})-(3\sqrt{y}+6)\)
\(=\sqrt{x}(\sqrt{y}+2)-3(\sqrt{y}+2)=(\sqrt{x}-3)(\sqrt{y}+2)\)