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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 \(x+1=0\Leftrightarrow x=-1\)
2) \(a.\sqrt{\left(-15\right)^2}=\left|15\right|=15\)
\(b.\sqrt{\left(2-\sqrt{5}\right)^2}=\left|2-\sqrt{5}\right|=2-\sqrt{5}\)
\(c.\sqrt{\left(3-\sqrt{7}\right)^2}=\left|3-\sqrt{7}\right|=3-\sqrt{7}\)
\(d.2\sqrt{a^2}=\left|2\sqrt{a}\right|=2\sqrt{a}\)
bai1
a)
√x=5=> x=25
b)
√(x+1)=2; x+1=4;x=3
c)
√(x^2)=4=> x=±4; x>0=>x=4
d)
2√(x+1)=4=>√(x+1)=2
x+1=4;x=3
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 \(S=\left\{8\right\}\)
Ta có : \(\sqrt{3}.x-\sqrt{75}=0\)
\(\Leftrightarrow\sqrt{3}.x-5\sqrt{3}=0\)
\(\Leftrightarrow\sqrt{3}\left(x-5\right)=0\)
Vì \(\sqrt{3}\ne0\)
Nên : x - 5 = 0
Vậy x = 5.
b) Ta có : \(\sqrt{2}.x+\sqrt{2}=\sqrt{8}+\sqrt{32}\)
\(\Leftrightarrow\sqrt{2}\left(x+1\right)=6\sqrt{2}\)
\(\Leftrightarrow\sqrt{2}\left(x+1\right)-6\sqrt{2}=0\)
\(\Leftrightarrow\sqrt{2}.\left(x+1-6\right)=0\)
\(\Leftrightarrow\sqrt{2}.\left(x-5\right)=0\)
Vì \(\sqrt{2}\ne0\)
Nên x - 5 = 0
Suy ra : x = 5
Lời giải :
a) \(A=3\sqrt{x-1}+7\ge7\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=1\)
b) \(B=\frac{4}{\sqrt{x}+3}\le\frac{4}{3}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=0\)
c) \(C=\frac{3\sqrt{x}+8}{\sqrt{x}+3}=\frac{3\left(\sqrt{x}+3\right)-1}{\sqrt{x}+3}=3-\frac{1}{\sqrt{x}+3}\)
Có \(\frac{1}{\sqrt{x}+3}\le\frac{1}{3}\forall x\)
\(\Leftrightarrow-\frac{1}{\sqrt{x}+3}\ge\frac{-1}{3}\)
\(\Leftrightarrow3-\frac{1}{\sqrt{x}+3}\ge3-\frac{1}{3}=\frac{8}{3}\)
\(\Leftrightarrow C\ge\frac{8}{3}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=0\)
d) \(D=x-3\sqrt{x}+2\)
\(D=\left(\sqrt{x}\right)^2-2\cdot\sqrt{x}\cdot\frac{3}{2}+\frac{9}{4}-\frac{1}{4}\)
\(D=\left(\sqrt{x}-\frac{3}{2}\right)^2-\frac{1}{4}\ge\frac{-1}{4}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}=\frac{3}{2}\Leftrightarrow x=\frac{9}{4}\)
e) \(E=\frac{4}{x-2\sqrt{x}+3}=\frac{4}{\left(\sqrt{x}-1\right)^2+2}\le\frac{4}{2}=2\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\)
a) Vì \(3\sqrt{x-1}\ge0\forall x\ge1\)
\(\Rightarrow3\sqrt{x-1}+7\ge7\forall x\ge1\)
Dấu "=" xảy ra <=>\(3\sqrt{x-1}=0\Leftrightarrow\sqrt{x-1}=0\Leftrightarrow x-1=0\Leftrightarrow x=1\)
Vậy Amin =7 tại x=1
a) ĐK: \(x\ge -1\)
Ta có: \(x^2+\sqrt{x+1}=1\)
\(\Leftrightarrow (x^2-1)+\sqrt{x+1}=0\)
\(\Leftrightarrow (x-1)(x+1)+\sqrt{x+1}=0\)
\(\Leftrightarrow \sqrt{x+1}[(x-1)\sqrt{x+1}+1]=0\)
\(\Rightarrow \left[\begin{matrix} \sqrt{x+1}=0(1)\\ (x-1)\sqrt{x+1}+1=0(2)\end{matrix}\right.\)
Với \((1)\Rightarrow x+1=0\Rightarrow x=-1\) (thỏa mãn)
Với \((2)\Rightarrow x\sqrt{x+1}-(\sqrt{x+1}-1)=0\)
\(\Leftrightarrow x\sqrt{x+1}-\frac{x}{\sqrt{x+1}+1}=0\)
\(\Leftrightarrow x\left(\sqrt{x+1}-\frac{1}{\sqrt{x+1}+1}\right)=0\)
\(\Leftrightarrow x.\frac{x+1+\sqrt{x+1}-1}{\sqrt{x+1}+1}=0\)
\(\Leftrightarrow x.\frac{x+\sqrt{x+1}}{\sqrt{x+1}+1}=0\)
\(\Rightarrow \left[\begin{matrix} x=0\\ x+\sqrt{x+1}=0\end{matrix}\right.\)
Với \(x+\sqrt{x+1}=0\Rightarrow x=-\sqrt{x+1}\Rightarrow \left\{\begin{matrix} x\leq 0\\ x^2=x+1\end{matrix}\right.\Rightarrow x=\frac{1-\sqrt{5}}{2}\)
Vậy \(x=\left\{-1; \frac{1-\sqrt{5}}{2}; 0\right\}\)
b) ĐK: \(-3\leq x\leq 6\)
Ta có: \((\sqrt{3+x}+\sqrt{6-x})^2=3+x+6-x+2\sqrt{(3+x)(6-x)}\)
\(=9+2\sqrt{(3+x)(6-x)}\geq 9\)
\(\Rightarrow \sqrt{3+x}+\sqrt{6-x}\geq 3\) do \(\sqrt{3+x}+\sqrt{6-x}\) không âm.
Dấu "=" xảy ra khi \(\sqrt{(3+x)(6-x)}=0\Leftrightarrow x=-3; x=6\)
Vậy \(x=-3\) or $x=6$
a) \(\sqrt{x}< 3\)<=> x<9
b)\(\sqrt{4-x}\) ≤ 2 <=> 4 - x ≤ 4 <=> x≥0
c)\(\sqrt{x+2}=\sqrt{4-x}\) <=> x+2=4-x <=>2x=2<=>x=1
Vậy x=1
d)\(\sqrt{x^2-1}\)=x-1 <=> x\(^2\)-1=x\(^2\)-2x+1 <=> x\(^2\)-\(x^2\)-2x+1+1=0 <=> 2x=2 <=> x=1
Vậy x=1
Câu b á 0≤x≤4 nha