1. So sánh \(\sqrt{2011}\)+ \(\sqrt{2013}và2.\sqrt{2012}\)
2.Giải phương trình \(\sqrt{x+2.\sqrt{x-1}}+\sqrt{x-2.\sqrt{x-1}}=2\)
3. TÍnh giá trị biểu thức: A = \(\tan^230^o.\cos^230^o+2.\sin60^o+tan45^o-\tan60^o+\cos^230^o\)
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\(A=\sqrt{x^2-2x+1}+\sqrt{x^2+4x+4}\)
\(=\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+2\right)^2}\)
\(=|1-x|+|x+2|\ge|1-x+x+2|=3\)
\(x\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=2\)
\(\Leftrightarrow x\sqrt{\left(\sqrt{x+\frac{1}{4}}+\frac{1}{2}\right)^2}=2\)
\(\Leftrightarrow x\sqrt{x+\frac{1}{4}}+\frac{1}{2}=2\)
\(\Leftrightarrow x\sqrt{x+\frac{1}{4}}=\frac{3}{2}\)
Làm nốt
1) ĐKXĐ: \(\left\{{}\begin{matrix}x\ne0\\x+1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ge-1\end{matrix}\right.\)
2) \(A=2^2+\left(3\sqrt{2}\right)^2+2.2.3\sqrt{2}-12\sqrt{2}=4+18+12\sqrt{2}-12\sqrt{2}=22\)\(B=\sqrt{4+3+4\sqrt{3}-\sqrt{3}=\sqrt{7+3\sqrt{3}}}\)
3) a) \(A=\dfrac{x\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{2x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{x\sqrt{x}-2x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}\left(x-2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}=\sqrt{x}-1\)b) Ta có :
\(x=3+2\sqrt{2}=\left(\sqrt{2}\right)^2+2.1.\sqrt{2}+1^2=\left(\sqrt{2}+1\right)^2\)Thay x vào A ta đc : \(A=\sqrt{x}-1=\sqrt{\left(\sqrt{2}+1\right)^2}-1=\sqrt{2}+1-1=\sqrt{2}\)4) a)
\(\sqrt{9x-27}+\sqrt{x-3}-\dfrac{1}{2}\sqrt{4x-12}=7\Leftrightarrow3\sqrt{x-3}+\sqrt{x-3}-\dfrac{1}{2}.2.\sqrt{x-3}=7\Leftrightarrow3\sqrt{x-3}=7\Leftrightarrow x-3=\dfrac{49}{9}\Leftrightarrow x=\dfrac{76}{9}\)b)Đề chuyển thánh sinB=3/4 nha
Ta có: sin2B+cos2B=1=> cosB=\(\dfrac{\sqrt{7}}{4}\)
cosC=sinB=3/4
a,
\(\Leftrightarrow\sqrt{1-x}=\frac{x-1}{\sqrt{6-x}+\sqrt{-5-2x}}\)
\(\Leftrightarrow-\sqrt{1-x}=\sqrt{6-x}+\sqrt{-5-2x}\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{1-x}=\sqrt{6-x}-\sqrt{-5-2x}\\-\sqrt{1-x}=\sqrt{6-x}+\sqrt{-5-2x}\end{cases}}\)
b,tự nàm
c,
\(\Leftrightarrow64x^2-64x-64=64\sqrt{8x+1}\)
\(\Leftrightarrow\left(8x+1\right)^2=10\left(8x+1\right)+64\sqrt{8x+1}+55\)
đặt \(\sqrt{8x+1}=a\)
=>a4=10a2+64a+55
nhận thấy phương trình có dạng x4=ax2+bx+c
tìm số m sao cho b2-4(2m+a)(m2+c)=0
sau đó đưa về (x2+m)2=k2 với k là 1 số bất kì,sau đó giải ra
b)đk \(x\ge1\)
\(\sqrt{1+x^2+\frac{x^2}{\left(x+1\right)^2}}+\frac{x}{x+1}=\sqrt{\frac{\left(x+1\right)^2+x^2.\left(x+1\right)^2+x^2}{\left(x+1\right)^2}}+\frac{x}{x+1}\)
\(=\sqrt{\frac{x^4+2x^3+3x^2+2x+1}{\left(x+1\right)^2}}+\frac{x}{x+1}\)
\(=\sqrt{\frac{\left(x^2+x+1\right)^2}{\left(x+1\right)^2}}+\frac{x}{x+1}\)
\(=\frac{x^2+x+1}{x+1}+\frac{x}{x+1}=x+1\)
\(\Rightarrow\sqrt{1+2012^2+\frac{2012^2}{2013^2}}+\frac{2012}{2013}=2013\)
\(\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}=2013\)
\(\Leftrightarrow\left|x-1\right|+\left|x-2\right|=2013\)
\(\Leftrightarrow x+\left|x-2\right|=2014\)
giai 2 pt
pt1 x+x-2=2014
x=1008
pt2 x+2-x=2014(vô lý)
b) đk: \(x>2012;y>2013\)
pt \(\frac{16}{\sqrt{x-2012}}+\sqrt{x-2012}+\frac{1}{\sqrt{y-2013}}+\sqrt{y-2013}=10\)
\(VT\ge2\sqrt{\frac{16}{\sqrt{x-2012}}.\sqrt{x-2012}}+2\sqrt{\frac{1}{\sqrt{y-2013}}.\sqrt{y-2013}}=8+2=10\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}x-2012=16\\y-2013=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2028\\y=2014\end{cases}}\)
a: Sửa đề: \(B=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
Khi x=9 thì \(B=\dfrac{\sqrt{9}+1}{\sqrt{9}+2}\)
\(=\dfrac{3+1}{3+2}=\dfrac{4}{5}\)
b: \(A=\dfrac{\sqrt{x}-3}{\sqrt{x}+2}+\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{6+\sqrt{x}}{x-4}\)
\(=\dfrac{\sqrt{x}-3}{\sqrt{x}+2}+\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{\sqrt{x}+6}{\left(\sqrt{x}-2\right)\cdot\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)+\sqrt{x}\left(\sqrt{x}+2\right)-\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x-5\sqrt{x}+6+x+2\sqrt{x}-\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2x-4\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{2\sqrt{x}}{\sqrt{x}+2}\)
c: P=A/B
\(=\dfrac{2\sqrt{x}}{\sqrt{x}+2}:\dfrac{\sqrt{x}+1}{\sqrt{x}+2}=\dfrac{2\sqrt{x}}{\sqrt{x}+1}\)
\(P-2=\dfrac{2\sqrt{x}}{\sqrt{x}+1}-2=\dfrac{2\sqrt{x}-2\sqrt{x}-2}{\sqrt{x}+1}\)
\(=\dfrac{-2}{\sqrt{x}+1}< 0\)
=>P<2
Câu 4:
Giả sử điều cần chứng minh là đúng
\(\Rightarrow x=y\), thay vào điều kiện ở đề bài, ta được:
\(\sqrt{x+2014}+\sqrt{2015-x}-\sqrt{2014-x}=\sqrt{x+2014}+\sqrt{2015-x}-\sqrt{2014-x}\) (luôn đúng)
Vậy điều cần chứng minh là đúng
2) \(\sqrt{x^2-5x+4}+2\sqrt{x+5}=2\sqrt{x-4}+\sqrt{x^2+4x-5}\)
⇔ \(\sqrt{\left(x-4\right)\left(x-1\right)}-2\sqrt{x-4}+2\sqrt{x+5}-\sqrt{\left(x+5\right)\left(x-1\right)}=0\)
⇔ \(\sqrt{x-4}.\left(\sqrt{x-1}-2\right)-\sqrt{x+5}\left(\sqrt{x-1}-2\right)=0\)
⇔ \(\left(\sqrt{x-4}-\sqrt{x+5}\right)\left(\sqrt{x-1}-2\right)=0\)
⇔ \(\left[{}\begin{matrix}\sqrt{x-4}-\sqrt{x+5}=0\\\sqrt{x-1}-2=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}\sqrt{x-4}=\sqrt{x+5}\\\sqrt{x-1}=2\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x\in\varnothing\\x=5\end{matrix}\right.\)
⇔ x = 5
Vậy S = {5}