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Do \(x^6-x^3+x^2-x+1=\left(x^3-\dfrac{1}{2}\right)^2+\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}>0\) ; \(\forall x\) nên BPT tương đương:
\(\sqrt{13}-\sqrt{2x^2-2x+5}-\sqrt{2x^2-4x+4}\ge0\)
\(\Leftrightarrow\sqrt{4x^2-4x+10}+\sqrt{4x^2-8x+8}\le\sqrt{26}\) (1)
Ta có:
\(VT=\sqrt{\left(2x-1\right)^2+3^2}+\sqrt{\left(2-2x\right)^2+2^2}\ge\sqrt{\left(2x-1+2-2x\right)^2+\left(3+2\right)^2}=\sqrt{26}\) (2)
\(\Rightarrow\left(1\right);\left(2\right)\Rightarrow\sqrt{4x^2-4x+10}+\sqrt{4x^2-8x+8}=\sqrt{26}\)
Dấu "=" xảy ra khi và chỉ khi \(2\left(2x-1\right)=3\left(2-2x\right)\Leftrightarrow x=\dfrac{4}{5}\)
Vậy BPT có nghiệm duy nhất \(x=\dfrac{4}{5}\)
Bạn xem lại đề câu b và c nhé !
a) \(\sqrt{x^2+2x+4}\ge x-2\) \(\left(ĐK:x\ge2\right)\)
\(\Leftrightarrow x^2+2x+4>x^2-4x+4\)
\(\Leftrightarrow6x>0\Leftrightarrow x>0\) kết hợp với ĐKXĐ
\(\Rightarrow x\ge2\) thỏa mãn đề.
d) \(x+y+z+4=2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{z-5}\)
\(ĐKXĐ:x\ge2,y\ge3,z\ge5\)
Pt tương đương :
\(\left(x-2-2\sqrt{x-2}+1\right)+\left(y-3-4\sqrt{y-3}+4\right)+\left(z-5-6\sqrt{z-5}+9\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-5}-3\right)^2=0\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\sqrt{x-2}=1\\\sqrt{y-3}=2\\\sqrt{z-5}=3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=3\\y=7\\z=14\end{cases}}\) ( Thỏa mãn ĐKXĐ )
e) \(\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}=\frac{1}{2}\left(x+y+z\right)\) (1)
\(ĐKXĐ:x\ge0,y\ge1,z\ge2\)
Phương trình (1) tương đương :
\(x+y+z-2\sqrt{x}-2\sqrt{y-1}-2\sqrt{z-2}=0\)
\(\Leftrightarrow\left(x-2\sqrt{x}+1\right)+\left(y-1-2\sqrt{y-1}+1\right)+\left(z-2-2\sqrt{z-2}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-1}-1\right)^2+\left(\sqrt{z-2}-1\right)^2=0\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\sqrt{x}=1\\\sqrt{y-1}=1\\\sqrt{z-2}=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=2\\z=3\end{cases}}\)( Thỏa mãn ĐKXĐ )
\(Xét-mẫu-của-biểu-thức:\left(đk:x\ge1\right).ta-có:x-\sqrt{2\left(x^2+5\right)}=\frac{-\left(x^2+10\right)}{x+\sqrt{2\left(x^2+5\right)}}< 0\\
.\)Vậy nó luôn <0 với đk x>=1
\(Xét-tử:đặt-nó-bằng-A=\left(x-2\right)^2-\left(\sqrt{x-1}-1\right)^2\left(2x-1\right)=2\sqrt{x-1}\left(2x-1\right)-\left(x-1\right)\left(x+4\right)\\ =\sqrt{x-1}\left(2\left(2x-1\right)-\sqrt{x-1\left(x+4\right)}\right)\ge0.\\ \)\(=>\left(2\left(2x-1\right)-\sqrt{\left(x-1\right)}\left(x+4\right)\right)\ge0< =>\frac{\left(5-x\right)\left(x-2\right)^2}{2\left(2x-1\right)+\left(x-1\right)\left(x+4\right)}\ge0< =>x\le5\) Vậy . \(1\le x\le5\)
bài 2
ta có \(\left(\sqrt{8a^2+1}+\sqrt{8b^2+1}+\sqrt{8c^2+1}\right)^2\)
\(=\left(\sqrt{a}.\sqrt{\frac{8a^2+1}{a}}+\sqrt{b}.\sqrt{\frac{8b^2+1}{b}}+\sqrt{c}.\sqrt{\frac{8c^2+1}{c}}\right)^2\)\(=\left(A\right)\)
Áp dụng bất đẳng thức Bunhiacopxki ta có;
\(\left(A\right)\le\left(a+b+c\right)\left(8a+\frac{1}{a}+8b+\frac{1}{b}+8c+\frac{8}{c}\right)\)
\(=\left(a+b+c\right)\left(9a+9b+9c\right)=9\left(a+b+c\right)^2\)
\(\Rightarrow3\left(a+b+c\right)\ge\sqrt{8a^2+1}+\sqrt{8b^2+1}+\sqrt{8c^2+1}\)(đpcm)
Dấu \(=\)xảy ra khi \(a=b=c=1\)
bài 1:
a:\(\sqrt{\left(\sqrt{3}-2\right)^2}\)+\(\sqrt{\left(1+\sqrt{3}\right)^2}\)
=\(\sqrt{3}-2+1+\sqrt{3}\)
=\(2\sqrt{3}-1\)
b; dài quá mink lười làm thông cảm
bài 2:
\(\sqrt{x^2-2x+1}=7\)
=>\(\sqrt{\left(x-1\right)^2}=7
\)
=>\(\orbr{\begin{cases}x-1=7\\x-1=-7\end{cases}}\)
=>\(\orbr{\begin{cases}x=8\\x=-6\end{cases}}\)
b: \(\sqrt{4x-20}-3\sqrt{\frac{x-5}{9}}=\sqrt{1-x}\)
=>\(\sqrt{4\left(x-5\right)}-9\sqrt{x-5}=\sqrt{1-x}\)
\(=2\sqrt{x-5}-9\sqrt{x-5}=\sqrt{1-x}\)
=>\(-7\sqrt{x-5}=\sqrt{1-x}\)
=\(-7.\left(x-5\right)=1-x\)
=>\(-7x+35=1-x\)
=>\(-7x+x=1-35\)
=>\(-6x=-34\)
=>\(x\approx5.667\)
mink sợ câu b bài 2 sai đó bạn
1 a)\(\sqrt{\left(\sqrt{3}-2\right)^2}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
= \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
= \(|2-\sqrt{3}|+|1+\sqrt{3}|\)
= \(2-\sqrt{3}+1+\sqrt{3}\)
= \(2+1\)= \(3\)
b) \(\left(\frac{3}{2}\sqrt{6}+2\sqrt{\frac{2}{3}}-4\sqrt{\frac{3}{2}}\right)\cdot\left(3\sqrt{\frac{2}{3}}-\sqrt{12}-\sqrt{6}\right)\)
= \(\left(\frac{3}{2}\sqrt{6}+2\sqrt{\frac{6}{3^2}}-4\sqrt{\frac{6}{2^2}}\right)\cdot\left(3\sqrt{\frac{6}{3^2}}-\sqrt{6}\sqrt{2}-\sqrt{6}\right)\)
= \(\left(\frac{3}{2}\sqrt{6}+\frac{2}{3}\sqrt{6}-\frac{4}{2}\sqrt{6}\right)\cdot\left(\frac{3}{3}\sqrt{6}-\sqrt{6}\cdot\sqrt{2}-\sqrt{6}\right)\)
= \(\left(\frac{3}{2}\sqrt{6}+\frac{2}{3}\sqrt{6}-2\sqrt{6}\right)\cdot\left(\sqrt{6}-\sqrt{6}\cdot\sqrt{2}-\sqrt{6}\right)\)
= \(\left(\sqrt{6}\left(\frac{3}{2}+\frac{2}{3}-2\right)\right)\cdot\left(\sqrt{6}\left(1-\sqrt{2}-1\right)\right)\)
= \(\sqrt{6}\frac{1}{6}\cdot\sqrt{6}\left(-\sqrt{2}\right)\)
= \(\sqrt{6}^2\left(\frac{-\sqrt{2}}{6}\right)\)
= \(6\frac{-\sqrt{2}}{6}\)=\(-\sqrt{2}\)
2 a) \(\sqrt{x^2-2x+1}=7\)
<=> \(\sqrt{x^2-2x\cdot1+1^2}=7\)
<=> \(\sqrt{\left(x-1\right)^2}=7\)
<=> \(|x-1|=7\)
Nếu \(x-1>=0\)=>\(x>=1\)
=> \(|x-1|=x-1\)
\(x-1=7\)<=>\(x=8\)(thỏa)
Nếu \(x-1< 0\)=>\(x< 1\)
=> \(|x-1|=-\left(x-1\right)=1-x\)
\(1-x=7\)<=>\(-x=6\)<=> \(x=-6\)(thỏa)
Vậy x=8 hoặc x=-6
b) \(\sqrt{4x-20}-3\sqrt{\frac{x-5}{9}}=\sqrt{1-x}\)
<=> \(\sqrt{4\left(x-5\right)}-3\frac{\sqrt{x-5}}{3}=\sqrt{1-x}\)
<=> \(2\sqrt{x-5}-\sqrt{x-5}=\sqrt{1-x}\)
<=> \(\sqrt{x-5}=\sqrt{1-x}\)
ĐK \(x-5>=0\)<=> \(x=5\)
\(1-x\)<=> \(-x=-1\)<=> \(x=1\)
Ta có \(\sqrt{x-5}=\sqrt{1-x}\)
<=> \(\left(\sqrt{x-5}\right)^2=\left(\sqrt{1-x}\right)^2\)
<=> \(x-5=1-x\)
<=> \(x-x=1+5\)
<=> \(0x=6\)(vô nghiệm)
Vậy phương trình vô nghiệm
Kết bạn với mình nha :)
