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Đặt \(\sqrt[3]{1-x}=a;\text{ }\sqrt[3]{1+x}=b\Rightarrow a^3+b^3=2\)
Pt đã cho trở thành \(a+b=1\Leftrightarrow b=1-a\)
Suy ra: \(a^3+\left(1-a\right)^3=2\Leftrightarrow3a^2-3a-1=0\Leftrightarrow a=\frac{3\pm\sqrt{21}}{6}\)
\(\Leftrightarrow\sqrt[3]{1-x}=\frac{3\pm\sqrt{21}}{6}\Leftrightarrow x=1-\left(\frac{3\pm\sqrt{21}}{6}\right)^3\)
ĐKXĐ: \(x>2;y>1\)
Khi đó Pt \(\Leftrightarrow\)\(\frac{36}{\sqrt{x-2}}+4\sqrt{x-2}+\frac{4}{\sqrt{y-1}}+\sqrt{y-1}=28\)
theo BĐT Cô si ta có \(\frac{36}{\sqrt{x-2}}+4\sqrt{x-2}\ge2.\sqrt{\frac{36}{\sqrt{x-2}}.4\sqrt{x-2}=24}\)
và \(\frac{4}{\sqrt{y-1}}+\sqrt{y-1}\ge2\sqrt{\frac{4}{\sqrt{y-1}}.\sqrt{y-1}}=4\)
Pt đã cho có VT>= 28 Dấu "=" xảy ra \(\Leftrightarrow\)
\(\frac{36}{\sqrt{x-2}}=4\sqrt{x-2}\Leftrightarrow x=11\)
và \(\frac{4}{\sqrt{y-1}}=\sqrt{y-1}\Leftrightarrow y=5\)
Đối chiếu với ĐK thì x=11; y=5 là nghiệm của PT
1.
đặt \(a=\sqrt{2+\sqrt{x}}\),\(b=\sqrt{2-\sqrt{x}}\)\(\left(a,b>0\right)\)
có \(a^2+b^2=4\)
pt thành \(\frac{a^2}{\sqrt{2}+a}+\frac{b^2}{\sqrt{2}-b}=\sqrt{2}\)
\(\Leftrightarrow\sqrt{2}\left(a^2+b^2\right)-ab\left(a-b\right)=\sqrt{2}\left(\sqrt{2}+a\right)\left(\sqrt{2}-b\right)\)
\(\Leftrightarrow2\sqrt{2}+\sqrt{2}ab-ab\left(a-b\right)-2\left(a-b\right)=0\)
\(\Leftrightarrow\left(ab+2\right)\left(\sqrt{2}-a+b\right)=0\)
vì a,b>o nên \(a-b=\sqrt{2}\)
\(\Rightarrow\sqrt{2+\sqrt{x}}-\sqrt{2-\sqrt{x}}=\sqrt{2}\)
Bình phương 2 vế:
\(4-2\sqrt{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}=2\)
\(\Leftrightarrow\sqrt{4-x}=1\)
\(\Rightarrow x=3\)
1) Sửa đề: \(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
Ta có: \(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{15\sqrt{x}-11-\left(3x+7\sqrt{x}-6\right)-\left(2x+\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{-5x+5\sqrt{x}+2\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{-5\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(-5\sqrt{x}+2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
Ta có: \(x=3-2\sqrt{2}\)
\(=2-2\cdot\sqrt{2}\cdot1+1\)
\(=\left(\sqrt{2}-1\right)^2\)
Thay \(x=\left(\sqrt{2}-1\right)^2\) vào biểu thức \(A=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\), ta được:
\(A=\frac{-5\cdot\sqrt{\left(\sqrt{2}-1\right)^2}+2}{\sqrt{\left(\sqrt{2}-1\right)^2}+3}\)
\(=\frac{-5\cdot\left(\sqrt{2}-1\right)+2}{\sqrt{2}-1+3}\)
\(=\frac{-5\sqrt{2}+5+2}{\sqrt{2}+2}\)
\(=\frac{-5\sqrt{2}+7}{\sqrt{2}+2}\)
Vậy: Khi \(x=3-2\sqrt{2}\) thì \(A=\frac{-5\sqrt{2}+7}{\sqrt{2}+2}\)
2) Ta có: \(B=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+1}{x-1}\)
\(=\frac{\left(x+2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x\sqrt{x}+x+2\sqrt{x}+2+x+x\sqrt{x}-\sqrt{x}-1-\left(2x+2\sqrt{x}+x\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{2x+2x\sqrt{x}+\sqrt{x}+1-2x-2\sqrt{x}-x\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x\sqrt{x}-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}\left(x-1\right)}{\left(x-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
Ta có: \(x=7-2\sqrt{6}\)
\(=6-2\sqrt{6}\cdot1+1\)
\(=\left(\sqrt{6}-1\right)^2\)
Thay \(x=\left(\sqrt{6}-1\right)^2\) vào biểu thức \(B=\frac{\sqrt{x}}{x+\sqrt{x}+1}\), ta được:
\(B=\frac{\sqrt{\left(\sqrt{6}-1\right)^2}}{\left(\sqrt{6}-1\right)^2+\sqrt{\left(\sqrt{6}-1\right)^2}+1}\)
\(=\frac{\sqrt{6}-1}{7-2\sqrt{6}+\sqrt{6}-1+1}\)
\(=\frac{\sqrt{6}-1}{7-\sqrt{6}}\)
Vậy: Khi \(x=7-2\sqrt{6}\) thì \(B=\frac{\sqrt{6}-1}{7-\sqrt{6}}\)
3) Ta có: \(C=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)
\(=\left(\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\frac{x+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{x-3\sqrt{x}-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{x-3\sqrt{x}-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}\)
\(=\frac{\sqrt{x}\left(x-3\sqrt{x}-x-9\right)}{\left(\sqrt{x}+3\right)\left(2\sqrt{x}+4\right)}\)
\(=\frac{\sqrt{x}\left(-3\sqrt{x}-9\right)}{\left(\sqrt{x}+3\right)\cdot2\cdot\left(\sqrt{x}+2\right)}\)
\(=\frac{-3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(2\sqrt{x}+4\right)}\)
\(=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\)
Ta có: \(x=7-4\sqrt{3}\)
\(=4-2\cdot2\cdot\sqrt{3}+3\)
\(=\left(2-\sqrt{3}\right)^2\)
Thay \(x=\left(2-\sqrt{3}\right)^2\) vào biểu thức \(C=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\), ta được:
\(C=\frac{-3\cdot\sqrt{\left(2-\sqrt{3}\right)^2}}{2\cdot\sqrt{\left(2-\sqrt{3}\right)^2}+4}\)
\(=\frac{-3\cdot\left(2-\sqrt{3}\right)}{2\cdot\left(2-\sqrt{3}\right)+4}\)
\(=\frac{-6+3\sqrt{3}}{4-2\sqrt{3}+4}\)
\(=\frac{-6+3\sqrt{3}}{8-2\sqrt{3}}\)
Vậy: Khi \(x=7-4\sqrt{3}\) thì \(C=\frac{-6+3\sqrt{3}}{8-2\sqrt{3}}\)
từ dòng cuối là sai rồi bạn à
Bạn bỏ dòng cuối đi còn lại đúng rồi
Ở tử đặt nhân tử chung căn x chung rồi lại đặt căn x +1 chung
Ở mẫu tách 3 căn x ra 2 căn x +căn x rồi đặt nhân tử 2 căn x ra
rút gọn được \(\frac{3\sqrt{x}-5}{2\sqrt{x}+1}\)
ĐK: \(x\ge-1\)
\(\frac{pt\Leftrightarrow\sqrt{x+1}\sqrt{x^2-x+1}}{\sqrt{x+3}}+\sqrt{x+1}=\sqrt{x^2-x+1}+\sqrt{x+3}\)
\(\Leftrightarrow\frac{\sqrt{x+1}}{\sqrt{x+3}}\left(\sqrt{x^2-x+1}+\sqrt{x+3}\right)=\sqrt{x^2-x+1}+\sqrt{x+3}\)
\(\Leftrightarrow\frac{\sqrt{x+1}}{\sqrt{x+3}}=1\text{ (do }\sqrt{x^2-x+1}>0\text{)}\)
\(\Leftrightarrow...\)