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\(a,\sqrt{3-x}+\sqrt{2-x}=1\)
\(\Rightarrow\sqrt{3+x}=1-\sqrt{2-x}\)
\(\Rightarrow3+x=1-2\sqrt{2-x}+2-x\)
\(\Rightarrow2x+2\sqrt{2-x}=0\)
\(\Rightarrow x+\sqrt{2-x}=0\)
\(\Rightarrow2-x=\left(-x\right)^2\)
\(\Rightarrow2-x=x^2\)
\(\Rightarrow2-x^2-x=0\)
\(\Rightarrow x^2+x-2=0\)
\(\Rightarrow\orbr{\begin{cases}x+2=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}}}\)
Vậy....
\(a,\sqrt{x+1}=\sqrt{2-x}\)
\(\Rightarrow x+1=2-x\)
\(\Rightarrow2x=1\)
\(\Rightarrow x=\frac{1}{2}\)
a) \(ĐKXĐ:-1\le x\le2\)
Bình phương 2 vế ta có:
\(x+1=2-x\)\(\Leftrightarrow2x=1\)\(\Leftrightarrow x=\frac{1}{2}\)( đpcm )
Vậy \(x=\frac{1}{2}\)
b) \(ĐKXĐ:x\ge1\)
\(\sqrt{36x-36}-\sqrt{9x-9}-\sqrt{4x-4}=16-\sqrt{x-1}\)
\(\Leftrightarrow\sqrt{36\left(x-1\right)}-\sqrt{9\left(x-1\right)}-\sqrt{4\left(x-1\right)}+\sqrt{x-1}=16\)
\(\Leftrightarrow6\sqrt{x-1}-3\sqrt{x-1}-2\sqrt{x-1}+\sqrt{x-1}=16\)
\(\Leftrightarrow2\sqrt{x-1}=16\)\(\Leftrightarrow\sqrt{x-1}=8\)
\(\Leftrightarrow x-1=64\)\(\Leftrightarrow x=65\)( thỏa mãn ĐKXĐ )
Vậy \(x=65\)
c) \(ĐKXĐ:x\ge1\)
\(\sqrt{16x-16}-\sqrt{9x-9}+\sqrt{4x-4}+\sqrt{x-1}=8\)
\(\Leftrightarrow\sqrt{16\left(x-1\right)}-\sqrt{9\left(x-1\right)}+\sqrt{4\left(x-1\right)}+\sqrt{x-1}=8\)
\(\Leftrightarrow4\sqrt{x-1}-3\sqrt{x-1}+2\sqrt{x-1}+\sqrt{x-1}=8\)
\(\Leftrightarrow4\sqrt{x-1}=8\)\(\Leftrightarrow\sqrt{x-1}=2\)
\(\Leftrightarrow x-1=4\)\(\Leftrightarrow x=5\)( thỏa mãn ĐKXĐ )
Vậy \(x=5\)
a) \(\sqrt{x}+\sqrt{\frac{x}{9}}-\frac{1}{3}\sqrt{4x}=5\)
ĐK : x ≥ 0
<=>\(\sqrt{x}+\sqrt{x\times\frac{1}{9}}-\frac{1}{3}\sqrt{2^2x}=5\)
<=> \(\sqrt{x}+\sqrt{x\times\left(\frac{1}{3}\right)^2}-\left(\frac{1}{3}\times\left|2\right|\right)\sqrt{x}=5\)
<=> \(\sqrt{x}+\left|\frac{1}{3}\right|\sqrt{x}-\left(\frac{1}{3}\times2\right)\sqrt{x}=5\)
<=> \(\sqrt{x}+\frac{1}{3}\sqrt{x}-\frac{2}{3}\sqrt{x}=5\)
<=> \(\sqrt{x}\left(1+\frac{1}{3}-\frac{2}{3}\right)=5\)
<=> \(\sqrt{x}\times\frac{2}{3}=5\)
<=> \(\sqrt{x}=\frac{15}{2}\)
<=> \(x=\frac{225}{4}\)( tm )
a) \(3\sqrt{x}-2\sqrt{9x}+\sqrt{16x}=5\)
\(\Leftrightarrow3\sqrt{x}-6\sqrt{x}+4\sqrt{x}=5\)
\(\Leftrightarrow\sqrt{x}=5\)
<=> x = 25
b) pt <=> \(\left(x^2+5\right)=\left(x+1\right)^2\)
<=> \(\left(x^2+5\right)=x^2+2x+1\)
<=> 2x = 4
<=> x = 2
c) pt <=> \(45-14\sqrt{x}+x=x-11\)
<=> \(45+11=14\sqrt{x}\)
<=> \(56=14\sqrt{x}\)
<=> \(4=\sqrt{x}\)
<=> x = 16
p/s : Cậu tự đặt điều kiện nhé
\(a,\sqrt{2x+5}=\sqrt{1-x}\)
\(\Rightarrow2x+5=1-x\)
\(2x+x=1-5\)
\(3x=-4\Leftrightarrow x=\frac{-4}{3}\)
Vậy \(S=\left\{-\frac{4}{3}\right\}\)thuộc tập nghiệm của pt trên
Mấy bài này dài vật vã ghê =)))))))))))))
1, a, \(\frac{3+4\sqrt{3}}{\sqrt{6}+\sqrt{2}-\sqrt{5}}\)
= \(\frac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{\left(\sqrt{6}+\sqrt{2}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}\)
=\(\frac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{\left(\sqrt{6}+\sqrt{2}\right)^2-5}\)
=\(\frac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{8+4\sqrt{3}-5}\)
= \(\frac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{3+4\sqrt{3}}\)
=\(\sqrt{6}+\sqrt{2}+\sqrt{5}\)
b, M = \(\frac{\sqrt{3}\left(x-1\right)}{\sqrt{x^2}-x+1}\)(ĐKXĐ: \(x\ge0\))
= \(\frac{\sqrt{3}\left(x-1\right)}{x-x+1}\)
= \(\sqrt{3}\left(x-1\right)\)
Thay x = \(2+\sqrt{3}\)(TMĐK) vào M ta có:
M = \(\sqrt{3}\left(2+\sqrt{3}-1\right)=\sqrt{3}\left(1+\sqrt{3}\right)=3+\sqrt{3}\)
Vậy với x = \(2+\sqrt{3}\)thì M = \(3+\sqrt{3}\)
2, Mình chỉ giải câu a thôi nhé:
\(\sqrt{1+b}+\sqrt{1+c}\ge2\sqrt{1+a}\)
\(\Leftrightarrow\left(\sqrt{1+b}+\sqrt{1+c}\right)^2\ge\left(2\sqrt{1+a}\right)^2\)
\(\Leftrightarrow1+b+2\sqrt{\left(1+b\right)\left(1+c\right)}+1+c\ge4\left(1+a\right)\)
\(\Leftrightarrow2+b+c+2\sqrt{\left(1+b\right)\left(1+c\right)}\ge4\left(1+a\right)\left(1\right)\)
Vì \(\left(\sqrt{1+b}-\sqrt{1+c}\right)^2\ge0\)
\(\Rightarrow2+b+c\ge2\sqrt{\left(1+b\right)\left(1+c\right)}\left(2\right)\)
Từ \(\left(1\right),\left(2\right)\Rightarrow4+2\left(b+c\right)+2\sqrt{\left(1+b\right)\left(1+c\right)}\ge4\left(1+a\right)+2\sqrt{\left(1+b\right)\left(1+c\right)}\)
\(\Leftrightarrow4+2\left(b+c\right)\ge4\left(1+a\right)\)
\(\Leftrightarrow4+2\left(b+c\right)\ge4+4a\)
\(\Leftrightarrow2\left(b+c\right)\ge4a\)
\(\Leftrightarrow b+c\ge2a\)
4*. Thật ra cái này mình xài làm trội, làm giảm là được mà
Đặt A = \(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{n}}\)
\(\frac{1}{2}A=\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}+....+\frac{1}{2\sqrt{n}}\)
\(\frac{1}{2}A=\frac{1}{\sqrt{2}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{3}}+....+\frac{1}{\sqrt{n}+\sqrt{n}}\)
Ta có: \(\frac{1}{\sqrt{2}+\sqrt{2}}>\frac{1}{\sqrt{3}+\sqrt{2}}\)
\(\frac{1}{\sqrt{3}+\sqrt{3}}>\frac{1}{\sqrt{4}+\sqrt{3}}\)
+ .........................................................
\(\frac{1}{\sqrt{n}+\sqrt{n}}>\frac{1}{\sqrt{n+1}+\sqrt{n}}\)
Cộng tất cả vào
\(\Rightarrow\frac{1}{\sqrt{2}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{3}}+...+\frac{1}{\sqrt{n}+\sqrt{n}}>\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+...+\frac{1}{\sqrt{n+1}+\sqrt{n}}\)\(\frac{1}{2}A>\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}+...+\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}\)
\(\frac{1}{2}A>\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{n+1}-\sqrt{n}\)
\(\frac{1}{2}A>\sqrt{n+1}-\sqrt{2}\)
\(A>2\sqrt{n+1}-2\sqrt{2}>2\sqrt{n+1}-3\)
\(A+1>2\sqrt{n+1}-3+1\)
\(A+1>2\sqrt{n+1}-2\)
\(A+1>2\left(\sqrt{n+1}-1\right)\)
Vậy ta có điều phải chứng minh.
c, \(\sqrt{9x-9}-2\sqrt{x-1}=8\left(đk:x\ge1\right)\)
\(< =>\sqrt{9\left(x-1\right)}-2\sqrt{x-1}=8\)
\(< =>\sqrt{9}.\sqrt{x-1}-2\sqrt{x-1}=8\)
\(< =>3\sqrt{x-1}-2\sqrt{x-1}=8\)
\(< =>\sqrt{x-1}=8< =>\sqrt{x-1}=\sqrt{8}^2=\left(-\sqrt{8}\right)^2\)
\(< =>\orbr{\begin{cases}x-1=8\\x-1=-8\end{cases}< =>\orbr{\begin{cases}x=9\left(tm\right)\\x=-7\left(ktm\right)\end{cases}}}\)
d, \(\sqrt{x-1}+\sqrt{9x-9}-\sqrt{4x-4}=4\left(đk:x\ge1\right)\)
\(< =>\sqrt{x-1}+\sqrt{9\left(x-1\right)}-\sqrt{4\left(x-1\right)}=4\)
\(< =>\sqrt{x-1}+\sqrt{9}.\sqrt{x-1}-\sqrt{4}.\sqrt{x-1}=4\)
\(< =>\sqrt{x-1}+3\sqrt{x-1}-2\sqrt{x-1}=4\)
\(< =>\sqrt{x-1}\left(1+3-2\right)=4< =>2\sqrt{x-1}=4\)
\(< =>\sqrt{x-1}=\frac{4}{2}=2=\sqrt{2}^2=\left(-\sqrt{2}\right)^2\)
\(< =>\orbr{\begin{cases}x-1=2\\x-1=-2\end{cases}< =>\orbr{\begin{cases}x=3\left(tm\right)\\x=-1\left(ktm\right)\end{cases}}}\)
a) \(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{1+2\sqrt{5}+\left(\sqrt{5}\right)^2}+\sqrt{1-2\sqrt{5}+\left(\sqrt{5}\right)^2}\)\(=\sqrt{\left(1+\sqrt{5}\right)^2}+\sqrt{\left(1-\sqrt{5}\right)^2}=1+\sqrt{5}-\left(1-\sqrt{5}\right)=1+\sqrt{5}-1+\sqrt{5}=2\sqrt{5}\)
a) \(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}+1+\sqrt{5}-1=2\sqrt{5}\)
b) \(\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}\)
\(=\sqrt{\left(2\sqrt{2}+\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)
\(=2\sqrt{2}+\sqrt{5}+2\sqrt{2}-\sqrt{5}=4\sqrt{2}\)
c) \(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)
\(=2\sqrt{2\sqrt{3}}-10\sqrt{2\sqrt{3}}+8\sqrt{2\sqrt{3}}=0\)
\(\sqrt{x+6-4\sqrt{x+2}}-\sqrt{9-4\sqrt{5}}=0\left(đk:x\ge-2\right)\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x+2}-2\right)^2}=\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(\Leftrightarrow\left|\sqrt{x+2}-2\right|=\left|\sqrt{5}-2\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+2}-2=\sqrt{5}-2\\\sqrt{x+2}-2=2-\sqrt{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=5\\x+2=21-8\sqrt{5}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=19-8\sqrt{5}\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{3;19-8\sqrt{5}\right\}\)