Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 2:Áp dụng BĐT AM-GM ta có:
\(\frac{1}{x}+\frac{1}{y}\ge2\sqrt{\frac{1}{xy}}\)
\(\frac{1}{y}+\frac{1}{z}\ge2\sqrt{\frac{1}{yz}}\)
\(\frac{1}{x}+\frac{1}{z}\ge2\sqrt{\frac{1}{xz}}\)
CỘng theo vế 3 BĐT trên có:
\(2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge2\left(\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\right)\)
Khi x=y=z
Ta có: \(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}\)
\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}\)
\(\frac{1}{\sqrt{3}}>\frac{1}{\sqrt{100}}\)
\(..........................\)
\(\frac{1}{\sqrt{99}}>\frac{1}{\sqrt{100}}\)
\(\frac{1}{\sqrt{100}}=\frac{1}{\sqrt{100}}\)
Cộng theo vế ta có:
\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{100}}>\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}=\frac{100}{10}=10\)
e/ \(\sqrt{x-2}+\sqrt{6-x}=\sqrt{x^2-8x+24}\)
\(\Leftrightarrow4+2\sqrt{\left(x-2\right)\left(6-x\right)}=x^2-8x+24\)
\(\Leftrightarrow2\sqrt{-x^2+8x-12}=x^2-8x+20\)
Đặt \(\sqrt{-x^2+8x-12}=a\left(a\ge0\right)\)thì pt thành
\(2a=-a^2+8\)
\(\Leftrightarrow a^2+2a-8=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=-4\left(l\right)\\a=2\end{cases}}\)
\(\Leftrightarrow\sqrt{-x^2+8x-12}=2\)
\(\Leftrightarrow-x^2+8x-12=4\)
\(\Leftrightarrow\left(x-4\right)^2=0\Leftrightarrow x=4\)
a/ \(4x^2+3x+3-4x\sqrt{x+3}-2\sqrt{2x-1}=0\)
\(\Leftrightarrow\left(4x^2-4x\sqrt{x+3}+x+3\right)+\left(2x-1-2\sqrt{2x-1}+1\right)=0\)
\(\Leftrightarrow\left(2x-\sqrt{x+3}\right)^2+\left(1-\sqrt{2x-1}\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}2x=\sqrt{x+3}\\1=\sqrt{2x-1}\end{cases}\Leftrightarrow}x=1\)
1.Ta co:
\(\text{ }\sqrt{5x^2+10x+9}=\sqrt{5\left(x+1\right)^2+4}\ge2\)
\(\sqrt{2x^2+4x+3}=\sqrt{2\left(x+1\right)^2+1}\ge1\)
\(\Rightarrow A=\sqrt{5x^2+10x+9}+\sqrt{2x^2+4x+3}\ge2+1=3\)
Dau '=' xay ra khi \(x=-1\)
Vay \(A_{min}=3\)khi \(x=-1\)
a) \(\sqrt{4x}=10\) (ĐKXĐ: 4x>=0 <=> x>=0)
\(\Leftrightarrow4x=100\)
\(\Leftrightarrow x=25\)
\(S=\left\{25\right\}\)
b) \(\sqrt{x^2-2x+1}=8\)
\(\Leftrightarrow\sqrt{\left(x-1\right)^2}=8\)
\(\Leftrightarrow x-1=8\)
\(\Leftrightarrow x=9\)
\(S=\left\{9\right\}\)
c) \(\sqrt{x^2-6x+9}=\sqrt{1-6x+9x^2}\)
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=\sqrt{\left(1-3x\right)^2}\)
\(\Leftrightarrow x-3=1-3x\) hoặc \(\Leftrightarrow x-3=-1+3x\)
\(\Leftrightarrow x+3x=1+3\) \(\Leftrightarrow x-3x=-1+3\)
\(\Leftrightarrow4x=4\) \(\Leftrightarrow-2x=2\)
\(\Leftrightarrow x=1\) \(\Leftrightarrow x=-1\)
\(S=\left\{1;-1\right\}\)
d) \(\sqrt{2x-5}=x-2\)
\(\Leftrightarrow2x-5=x^2-4x+4\)
\(\Leftrightarrow-x^2+2x+4x-5-4=0\)
\(\Leftrightarrow-x^2+6x-9=0\)
\(\Leftrightarrow x^2-6x+9=0\)
\(\Leftrightarrow\left(x-3\right)^2=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
\(S=\left\{3\right\}\)
e) \(\sqrt{x^2-2x+1}=\sqrt{x+1}\)
\(\Leftrightarrow x^2-2x+1=x+1\)
\(\Leftrightarrow x^2-2x-x+1-1=0\)
\(\Leftrightarrow x^2-3x=0\)
\(\Leftrightarrow x\left(x-3\right)=0\)
\(\Leftrightarrow x=0\) hoặc \(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
\(S=\left\{0;3\right\}\)
g) \(\sqrt{x^2-9}-\sqrt{x-3}=0\) ( ĐKXĐ: x-3>=0 <=> x>=3)
\(\Leftrightarrow\sqrt{x^2-9}=\sqrt{x-3}\)
\(\Leftrightarrow x^2-9=x-3\)
\(\Leftrightarrow x^2-x-6=0\)
\(\Leftrightarrow x^2-3x+2x-6=0\)
\(\Leftrightarrow\left(x^2+2x\right)-\left(3x+6\right)=0\)
\(\Leftrightarrow x\left(x+2\right)-3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow x+2=0\) hoặc \(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=-2\) \(\Leftrightarrow x=3\)
\(S=\left\{-2;3\right\}\)
h) \(\sqrt{x^2-4x+4}+\sqrt{x^2-6x+9}=1\)
\(\Leftrightarrow\sqrt{\left(x-2\right)^2}+\sqrt{\left(x-3\right)^2}=1\)
\(\Leftrightarrow x-2+x-3-1=0\)
\(\Leftrightarrow2x-6=0\)
\(\Leftrightarrow x=3\)
\(S=\left\{3\right\}\)
i) \(\sqrt{\frac{2x-3}{x-1}}=2\)
\(\Leftrightarrow\frac{2x-3}{x-1}=4\)
\(\Leftrightarrow4\left(x-1\right)=2x-3\)
\(\Leftrightarrow4x-4-2x+3=0\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow x=\frac{1}{2}\)
\(S=\left\{\frac{1}{2}\right\}\)
l) \(x+y+12=4\sqrt{x}+6\sqrt{y-1}\)
\(\Leftrightarrow x+y-4\sqrt{x}+12-6\sqrt{y-1}=0\)
\(\Leftrightarrow\left(x-4\sqrt{x}+4\right)+\left(y-1-6\sqrt{y-1}+9\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+\left(\sqrt{y-1}-3\right)^2=0\)
\(\Leftrightarrow\sqrt{x}-2=0\) hoặc \(\Leftrightarrow\sqrt{y-1}-3=0\)
\(\Leftrightarrow\sqrt{x}=2\) \(\Leftrightarrow\sqrt{y-1}=3\)
\(\Leftrightarrow x=4\) \(\Leftrightarrow y-1=9\)
\(\Leftrightarrow y=10\)
KẾT luận : ..............
Tới đây nhé, nếu mai chưa ai giải thì mình giải hộ cho
CHÚC BẠN HỌC TỐT!
m) \(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8+6\sqrt{x-1}}=5\)
<=> \(\sqrt{\left(x-1\right)-4\sqrt{x-1}+4}+\sqrt{\left(x-1\right)+6\sqrt{x-1}+9}=5\)
<=>\(\sqrt{\left(\sqrt{x-1}+2\right)^2}+\sqrt{\left(\sqrt{x-1}+3\right)^2}=5\)
<=>\(\sqrt{x-1}+2+\sqrt{x-1}+3=5\)
<=> \(2\sqrt{x-1}=0\)
<=> \(\sqrt{x-1}=0\) <=>x=1
Vậy \(S=\left\{1\right\}\)
n) \(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\) (*) ( đk \(x\ge\frac{1}{2}\))
<=> \(\left(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}\right)^2=2\)
<=> \(x+\sqrt{2x-1}+x-\sqrt{2x-1}+2\sqrt{x^2-2x+1}=2\)
<=> 2x+\(2\sqrt{\left(x-1\right)^2=2}\)
<=> x+\(\left|x-1\right|=2\)(1)
TH1: \(\frac{1}{2}\le x\le1\)
Từ (1) => x+1-x=2
<=> 1=2(vô lý)
TH2: x>1
Từ (1)=> x+x-1=2
<=> 2x=3<=> \(x=\frac{2}{3}\)(tm pt (*))
Vậy \(S=\left\{\frac{2}{3}\right\}\)
p) \(\sqrt{2x-1}+\sqrt{x-2}=\sqrt{x+1}\) (*) (đk :\(x\ge2\))
Đặt \(\left\{{}\begin{matrix}x-2=a\left(a\ge0\right)\\x+1=b\left(b\ge0\right)\end{matrix}\right.\) =>a+b=2x-1
Có \(\sqrt{a+b}+\sqrt{a}=\sqrt{b}\)
<=> \(\sqrt{a+b}=\sqrt{b}-\sqrt{a}\)
<=> \(a+b=b-2\sqrt{ab}+a\)
<=> 0=\(-2\sqrt{ab}\)
=> \(\left[{}\begin{matrix}a=0\\b=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\) => x=2 (vì x=-1 không thỏa mãn pt(*))
Vậy \(S=\left\{2\right\}\)
q) \(\sqrt{x-7}+\sqrt{9-x}=x^2-16x+66\)(*) (đk : \(7\le x\le9\))
Với a,b\(\ge0\) có: \(\sqrt{a}+\sqrt{b}\le2\sqrt{\frac{a+b}{2}}\)(tự cm nha) .Dấu "=" xảy ra <=> a=b
Áp dụng bđt trên có:
\(\sqrt{x-7}+\sqrt{9-x}\le2\sqrt{\frac{x-7+9-x}{2}}=2\sqrt{\frac{2}{2}}=2\) (1)
Có x2-16x+66=(x2-16x+64)+2=(x-8)2+2 \(\ge2\) với mọi x (2)
Từ (1),(2) .Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}x-7=9-x\\x-8=0\end{matrix}\right.\)<=>\(\left\{{}\begin{matrix}2x=16\\x=8\end{matrix}\right.\)<=>\(\left\{{}\begin{matrix}x=8\\x=8\end{matrix}\right.\)<=> x=8( tm pt (*))
Vậy \(S=\left\{8\right\}\)
a/ ĐK: \(x \ge -1\). Đặt \(\sqrt{x+1}=a \ge 0\)
PT: \(\Leftrightarrow6a-3a-2a=5\)
\(\Leftrightarrow a=5\)
\(\Leftrightarrow x+1=15\Leftrightarrow x=24\) (nhận)
b,c: Hai ý này đều làm theo cách bình phương hoặc đưa về phương trình chứa dấu giá trị tuyệt đối được nhé.
b) Cách 1: ĐKXĐ: Tự tìm
\(\sqrt{x^{2}-4x+4}=2\Leftrightarrow x^{2}-4x+4=4\Leftrightarrow x(x-4)=0\)
\(\Leftrightarrow x=0\) hoặc \(x=4\) cả 2 cái này đều TMĐK
Cách 2: \((\sqrt{x^2-4x+4}=2)\)
\(\Leftrightarrow \sqrt{(x-2)^2}=2\)
\(\Leftrightarrow \mid x-2\mid=2\)
Với \(x\geq 2\) thì :
\(x-2=2 \Leftrightarrow x=4\) (nhận)
Với \(x<2\) thì
\(-x-2=2\Leftrightarrow x=0\) (nhận)
Vậy \(S={0;4}\)
c) Cách 1: \(\sqrt{x^{2}-6x+9}=x-2\Leftrightarrow \left\{\begin{matrix}x\geq 2 \\ x^{2}-6x+9=x^{2}-4x+4 \end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix}x\geq 2 \\ x=\frac{5}{2} \end{matrix}\right.\)
Nghiệm TMĐK
Cách 2: \((\sqrt{x^2-6x+9}=x-2)\)
\(\Leftrightarrow \mid x-3\mid =x-2\)
Với \(x\geq 3\) thì
\(x-3=x-2\Leftrightarrow 0x=-1\) ( vô lý)
Với \(x<3\) thì
\(-x+3=x-2\Leftrightarrow -2x=-5 \Leftrightarrow x=\frac{5}{2}\)
Vậy \(S={\frac{5}{2}}\)
d) ĐKXĐ: Tự tìm
\(\sqrt{x^{2}+4}=\sqrt{2x+3}\Leftrightarrow x^{2}+4=2x+3\Leftrightarrow x^{2}-2x+1=0\Leftrightarrow (x-1)^{2}=0\)
\(\Leftrightarrow x=1\)
e) ĐKXĐ: \(x\geq \frac{3}{2}\)
\(\frac{\sqrt{2x-3}}{\sqrt{x-1}}=2\Leftrightarrow \frac{2x-3}{x-1}=4\Rightarrow 2x-3=4x-4\Leftrightarrow x=\frac{1}{2}\)
Nghiệm không TMĐK.
Phương trình vô nghiệm.
f) ĐKXĐ: \(x\geq \frac{-15}{2}\)
\(x+\sqrt{2x+15}=0\Leftrightarrow 2x+2\sqrt{2x+15}=0\Leftrightarrow 2x+15+2\sqrt{2x+15}+1-16=0\)
\(\Leftrightarrow (\sqrt{2x+15}+1)^{2}-4^{2}=0\Leftrightarrow (\sqrt{2x+15}+5)(\sqrt{2x+15}-3)=0\)
\(\Leftrightarrow \sqrt{2x+15}-3=0\Leftrightarrow \sqrt{2x+15}=3\Leftrightarrow 2x+15=9\Leftrightarrow x=-3\) (TMĐK)
1.a) \(\sqrt{x^2-4}-\sqrt{x-2}=0\)
\(\Leftrightarrow\sqrt{\left(x-2\right)\left(x+2\right)}-\sqrt{x-2}=0\)
\(\Leftrightarrow\sqrt{x-2}.\sqrt{x+2}-\sqrt{x-2}=0\)
\(\Leftrightarrow\sqrt{x-2}.\left(\sqrt{x+2}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x-2}=0\\\sqrt{x+2}-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\\sqrt{x+2}=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x+2=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)
Vậy x=2 hoặc x=-1
4.a)\(x-2\sqrt{x}+3\)
\(=x-2\sqrt{x}+1+2\)
\(=\left(\sqrt{x}-1\right)^2+2\)
Vì \(\left(\sqrt{x}-1\right)^2\ge0,\forall x\)
\(\left(\sqrt{x}-1\right)^2+2\ge2\)
\(\Rightarrow Min_{bt}=2\) khi \(\sqrt{x}-1=0\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\)
b)Ta có:
\(x-4\sqrt{y}+13\ge0\)
\(\Leftrightarrow x-4\sqrt{y}\ge-13\)
Dấu "=" xảy ra khi \(x-4\sqrt{y}=0\Leftrightarrow x=4\sqrt{y}\)
Vậy \(min_{bt}=0\) khi \(x=4\sqrt{y}\)
c)Ta có:
\(2x-4\sqrt{y}+6\ge0\)
\(\Leftrightarrow x-2\sqrt{y}+3\ge0\)
\(\Leftrightarrow x-2\sqrt{y}\ge-3\)
Dấu "=" xảy ra khi \(x-2\sqrt{y}=0\Leftrightarrow x=2\sqrt{y}\)
Vậy \(Min_{bt}=0\) khi \(x=2\sqrt{y}\)
d)Ta có:
\(x^2+2x+5=x^2+2x+1+4=\left(x+1\right)^2+4\)
Vì \(\left(x+1\right)^2\ge0,\forall x\)
\(\Leftrightarrow\left(x+1\right)^2+4\ge4\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)^2+4}\le\frac{1}{4}\)
\(\Leftrightarrow-\frac{1}{\left(x+1\right)^2+4}\ge-\frac{1}{4}\)
\(\Leftrightarrow-\frac{4}{\left(x+1\right)^2+4}\ge-1\)
Vậy \(Min_{bt}=-1\) khi \(x+1=0\Leftrightarrow x=-1\)
zài zậy