Giải phương trình:
a) \(\dfrac{1}{x+\sqrt{1+x^2}}+\dfrac{1}{x-\sqrt{1+x^2}}+2=0\)
b) \(2x-5a\sqrt{x-a}+2a\left(a-1\right)=0\) với a>0
* Giải phương trình
a. \(x^2-2\sqrt{5x}+5=0\)
b. \(\sqrt{x+3}=1\)
* Cho:
A=\(\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\) , với x>0 và x≠1
a. Rút gọn A
b. Tìm giá trị nhỏ nhất của A
Bài 1:
a: Ta có: \(x^2-2\sqrt{5}x+5=0\)
\(\Leftrightarrow x-\sqrt{5}=0\)
hay \(x=\sqrt{5}\)
b: Ta có: \(\sqrt{x+3}=1\)
\(\Leftrightarrow x+3=1\)
hay x=-2
giải phương trình :
a, \(2x^2-11x+21-3\sqrt[3]{4x-4}=0\)
b, \(\left(3x-2\right)\sqrt{x+1}-x^2-x-2=0\)
c, \(x+4-2\left(\dfrac{x+2}{x-1}\right)\sqrt{\dfrac{x-1}{x+2}}=0\)
c.
ĐKXĐ: \(\left[{}\begin{matrix}x>1\\x< -2\end{matrix}\right.\)
\(\Leftrightarrow x+4-2\sqrt[]{\left(\dfrac{x+2}{x-1}\right)^2\left(\dfrac{x-1}{x+2}\right)}=0\)
\(\Leftrightarrow x+4-2\sqrt[]{\dfrac{x+2}{x-1}}=0\)
\(\Leftrightarrow x+4=2\sqrt[]{\dfrac{x+2}{x-1}}\) (\(x\ge-4\))
\(\Leftrightarrow x^2+8x+16=\dfrac{4\left(x+2\right)}{x-1}\)
\(\Rightarrow x^3+7x^2+4x-24=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+4x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2+2\sqrt{3}\\x=-2-2\sqrt{3}\left(loại\right)\end{matrix}\right.\)
a.
\(\Leftrightarrow2x^2-11x+21=3\sqrt[3]{4\left(x-1\right)}\)
Do \(2x^2-11x+21=2\left(x-\dfrac{11}{4}\right)^2+\dfrac{47}{8}>0\Rightarrow3\sqrt[3]{4\left(x-1\right)}>0\Rightarrow x-1>0\)
Ta có:
\(VT=2x^2-11x+21-3\sqrt[3]{4x-4}=2\left(x^2-6x+9\right)+x+3-3\sqrt[3]{4\left(x-1\right)}\)
\(=2\left(x-3\right)^2+x+3-3\sqrt[3]{4\left(x-1\right)}\)
\(\Rightarrow VT\ge x+3-3\sqrt[3]{4\left(x-1\right)}=\left(x-1\right)+2+2-3\sqrt[3]{4\left(x-1\right)}\)
\(\Rightarrow VT\ge3\sqrt[3]{\left(x-1\right).2.2}-3\sqrt[3]{4\left(x-1\right)}=0\)
Đẳng thức xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}\left(x-3\right)^2=0\\x-1=2\\\end{matrix}\right.\) \(\Leftrightarrow x=3\)
Vậy pt có nghiệm duy nhất \(x=3\)
b.
ĐKXD: \(x\ge-1\)
Phương trình: \(2\left(x+1\right)-\left(3x-2\right)\sqrt[]{x+1}+x^2-x=0\)
Đặt \(\sqrt[]{x+1}=t\ge0\)
\(\Rightarrow2t^2-\left(3x-2\right)t+x^2-x=0\)
\(\Delta=\left(3x-2\right)^2-8\left(x^2-x\right)=\left(x-2\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{3x-2+x-2}{4}=x-1\\t=\dfrac{3x-2-x+2}{4}=\dfrac{x}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt[]{x+1}=x-1\left(x\ge1\right)\\\sqrt[]{x+1}=\dfrac{x}{2}\left(x\ge0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=x^2-2x+1\left(x\ge1\right)\\x+1=\dfrac{x^2}{4}\left(x\ge0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2+2\sqrt[]{2}\end{matrix}\right.\)
Tính giá trị của biểu thức
A=\(\dfrac{1+2x}{1+\sqrt{1+2x}}+\dfrac{1-2x}{1-\sqrt{1-2x}}\) với x=\(\dfrac{\sqrt{3}}{4}\)
B=\(\dfrac{2b\sqrt{x^2-1}}{x-\sqrt{x^2-1}}\) với x=\(\dfrac{1}{2}\left(\sqrt{\dfrac{a}{b}}+\sqrt{\dfrac{b}{a}}\right)\) và a>0,b>0
C=\(\dfrac{2a\sqrt{1+x^2}}{\sqrt{1+x^2}-x}\) với x=\(\dfrac{1}{2}\left(\sqrt{\dfrac{1-a}{a}}-\sqrt{\dfrac{a}{1-a}}\right)\) và 0<a<1
A)
Đặt \(\sqrt{1+2x}=a; \sqrt{1-2x}=b\) (\(a,b>0\) )
\(\Rightarrow \left\{\begin{matrix} a^2+b^2=2\\ a^2-b^2=4x=\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} 2a^2=2+\sqrt{3}\rightarrow 4a^2=4+2\sqrt{3}=(\sqrt{3}+1)^2\\ 2b^2=2-\sqrt{3}\rightarrow 4b^2=4-2\sqrt{3}=(\sqrt{3}-1)^2\end{matrix}\right.\)
\(\Rightarrow a=\frac{\sqrt{3}+1}{2}; b=\frac{\sqrt{3}-1}{2}\)
\(\Rightarrow ab=\frac{(\sqrt{3}+1)(\sqrt{3}-1)}{4}=\frac{1}{2}; a-b=1\)
Có:
\(A=\frac{a^2}{1+a}+\frac{b^2}{1-b}=\frac{a^2-a^2b+b^2+ab^2}{(1+a)(1-b)}\)
\(=\frac{2-ab(a-b)}{1+(a-b)-ab}=\frac{2-\frac{1}{2}.1}{1+1-\frac{1}{2}}=1\)
B)
\(2x=\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\)
\(\Rightarrow 4x^2=\frac{a}{b}+\frac{b}{a}+2\)
\(\rightarrow 4(x^2-1)=\frac{a}{b}+\frac{b}{a}-2=\left(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\right)^2\)
\(\Rightarrow \sqrt{4(x^2-1)}=\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\) do $a>b$
T có: \(B=\frac{b\sqrt{4(x^2-1)}}{x-\sqrt{x^2-1}}=\frac{2b\sqrt{4(x^2-1)}}{2x-\sqrt{4(x^2-1)}}=\frac{2b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}-\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}\)
\(=\frac{2b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{2\sqrt{\frac{b}{a}}}=\frac{b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{\sqrt{\frac{b}{a}}}=\frac{\frac{b(a-b)}{\sqrt{ab}}}{\sqrt{\frac{b}{a}}}=a-b\)
C)
\(2x=\sqrt{\frac{1-a}{a}}-\sqrt{\frac{a}{1-a}}\Rightarrow 4x^2=\frac{1-a}{a}+\frac{a}{1-a}-2\)
\(\Rightarrow 4(x^2+1)=\frac{1-a}{a}+\frac{a}{1-a}+2=(\sqrt{\frac{1-a}{a}}+\sqrt{\frac{a}{1-a}})^2\)
\(\Rightarrow \sqrt{4(x^2+1)}=\sqrt{\frac{1-a}{a}}+\sqrt{\frac{a}{1-a}}\)
Khi đó:
\(C=\frac{2a\sqrt{4(1+x^2)}}{\sqrt{4(x^2+1)}-2x}=\frac{2a\left ( \sqrt{\frac{1-a}{a}}+\sqrt{\frac{a}{1-a}} \right )}{\sqrt{\frac{1-a}{a}}+\sqrt{\frac{a}{1-a}}-(\sqrt{\frac{1-a}{a}}-\sqrt{\frac{a}{1-a}})}=\frac{a\left ( \sqrt{\frac{1-a}{a}}+\sqrt{\frac{a}{1-a}} \right )}{\sqrt{\frac{a}{1-a}}}\)
\(=\frac{\frac{a(1-a+a)}{\sqrt{a(1-a)}}}{\sqrt{\frac{a}{1-a}}}=1\)
Rút gọn :
B=\(\dfrac{2a\sqrt{1+x^2}}{\sqrt{1+x^2}-x}\) với x=\(\dfrac{1}{2}\left(\sqrt{\dfrac{1-a}{a}}-\sqrt{\dfrac{a}{1-a}}\right)\) và 0<a<1
giải các PT sau :
a) \(\left|2x+3\right|-\left|x\right|+\left|x-1\right|=2x+4\)
b) \(\sqrt{x}-\dfrac{4}{\sqrt{x+2}}+\sqrt{x+2}=0\)
c) \(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\)
d) \(x+\sqrt{x+\dfrac{1}{2}+\sqrt{x+\dfrac{1}{4}}}=4\)
e) \(\sqrt{4x+3}+\sqrt{2x+1}=6x+\sqrt{8x^2+10x+3}-16\)
f)\(\sqrt[3]{x-2}+\sqrt{x+1}=3\)
GIÚP MÌNH VỚI MÌNH ĐANG CẦN GẤP
bài 1: rut gọn
a, \(\sqrt{5\left\{1-a\right\}^2}\) với a>1
b,\(\sqrt{\dfrac{9\left[a^2+2a+1\right]}{144}}\)
c,\(\dfrac{2}{x-5}\times\sqrt{\dfrac{x^2\times10x+25}{64}}\)
d \(\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\) với x≥0 và x≠1
a: \(\sqrt{5\left(1-a\right)^2}\)
\(=\sqrt{5\left(a-1\right)^2}\)
\(=\sqrt{5}\cdot\sqrt{\left(a-1\right)^2}\)
\(=\sqrt{5}\left|a-1\right|\)
\(=\sqrt{5}\left(a-1\right)\)(do a>1 nên a-1>0)
b: \(\sqrt{\dfrac{9\left|a^2+2a+1\right|}{144}}\)
\(=\sqrt{\dfrac{9}{144}\cdot\left|a^2+2a+1\right|}\)
\(=\sqrt{\dfrac{1}{16}\cdot\left|\left(a+1\right)^2\right|}\)
\(=\sqrt{\dfrac{1}{16}}\cdot\sqrt{\left|\left(a+1\right)^2\right|}\)
\(=\dfrac{1}{4}\cdot\left(a+1\right)^2\)
c:
ĐKXĐ: x<>5
Sửa đề:\(\dfrac{2}{x-5}\cdot\sqrt{\dfrac{x^2-10x+25}{64}}\)
\(=\dfrac{2}{x-5}\cdot\sqrt{\dfrac{\left(x-5\right)^2}{64}}\)
\(=\dfrac{2}{x-5}\cdot\dfrac{\sqrt{\left(x-5\right)^2}}{\sqrt{64}}\)
\(=\dfrac{2}{x-5}\cdot\dfrac{\left|x-5\right|}{8}\)
\(=\pm\dfrac{1}{4}\)
d: \(\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\cdot\sqrt{x}-\sqrt{x}\cdot1}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=\sqrt{x}\)
b5: giải pt ;
a, \(\sqrt{49\left(1-2x+x^2\right)}-35=0\)
b, \(\sqrt{x^2-9}-5\sqrt{x+3}=0\)
c, \(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}+3}\)
a) Ta có: \(\sqrt{49\left(x^2-2x+1\right)}-35=0\)
\(\Leftrightarrow7\left|x-1\right|=35\)
\(\Leftrightarrow\left|x-1\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
b)
ĐKXĐ: \(\left[{}\begin{matrix}x\ge3\\x\le-3\end{matrix}\right.\)
Ta có: \(\sqrt{x^2-9}-5\sqrt{x+3}=0\)
\(\Leftrightarrow\sqrt{x+3}\left(\sqrt{x-3}-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+3}=0\\\sqrt{x-3}=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-3=25\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=28\left(nhận\right)\end{matrix}\right.\)
c) ĐKXĐ: \(x\ge0\)
Ta có: \(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}+3}\)
\(\Leftrightarrow x-1=x+\sqrt{x}-6\)
\(\Leftrightarrow\sqrt{x}-6=-1\)
\(\Leftrightarrow\sqrt{x}=5\)
hay x=25(nhận)
Cho biểu thức:
A = \(\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right)+\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)với x > 0; x ≠ 1
a) Chứng minh: A = \(\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b) Tìm x để 2A = \(2\sqrt{x}+5\)
a.
A = \(\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right)+\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
A = \(\left(\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\dfrac{1}{\sqrt{x}+2}\right)+\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
A = \(\left(\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\right)+\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
A = \(\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
A = \(\dfrac{\left(x-2+\sqrt{x}\right).\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
A = \(\dfrac{\left(x-2+\sqrt{x}\right)\left(\sqrt{x}-1\right)+\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
A = \(\dfrac{x\sqrt{x}-x-2\sqrt{x}+2+x-\left(x+\sqrt{x}\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
A = \(\dfrac{x\sqrt{x}-x-2\sqrt{x}+2+x-\left(x\sqrt{x}+2x+x+2\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
A = \(\dfrac{x\sqrt{x}-x-2\sqrt{x}+2+x-x\sqrt{x}-2x-x-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
A = \(\dfrac{-3x-4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
A = \(\dfrac{-\left(3x+4\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
A = \(\dfrac{-\sqrt{x}\left(3\sqrt{x}+4\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
A = \(\dfrac{3\sqrt{x}+4}{x-\sqrt{x}+2\sqrt{x}-2}\)
A = \(\dfrac{3\sqrt{x}+4}{x-\sqrt{x}-2}\)
a)A=\(\dfrac{1}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}\) với a>\(\dfrac{1}{2}\)
b)A=\(\dfrac{\sqrt{x-2\sqrt{x-1}}}{\sqrt{x-1}-1}\)+\(\dfrac{\sqrt{x+2\sqrt{x-1}}}{\sqrt{x-1+1}}\) với x>2
c)\(\dfrac{a+b}{b^2}\)\(\sqrt{\dfrac{a^2b^4}{a^2+2ab+b^2}}\) với a+b>0; b≠0
d)A=\(\left(\sqrt{\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\) với a≥0; a≠1
e)A=\(\dfrac{x-1}{\sqrt{y}-1}\sqrt{\dfrac{\left(y-2\sqrt{y}+1\right)}{\left(x-1\right)^4}}\) với x≠1; y≠1; y>o
f)A=\(\sqrt{\dfrac{m}{1-2x+x^2}}\)\(\sqrt{\dfrac{4m-8mx+4mx^2}{81}}\) với m>0; x≠4
g)A=\(\left(\dfrac{\sqrt{x}+1}{x-4}-\dfrac{\sqrt{x}-1}{x+4\sqrt{x}+4}\right)\)\(\dfrac{x\sqrt{x}+2x-4\sqrt{x}-8}{\sqrt{x}}\) với x>0; x≠4
h)\(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\)\(\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\) với a≥0; a≠1
a: \(A=\dfrac{1}{2a-1}\cdot\sqrt{5a^2}\cdot\left|2a-1\right|\)
\(=\dfrac{2a-1}{2a-1}\cdot a\sqrt{5}=a\sqrt{5}\)(do a>1/2)
b: \(A=\dfrac{\sqrt{x-1-2\sqrt{x-1}+1}}{\sqrt{x-1}-1}+\dfrac{\sqrt{x-1+2\sqrt{x-1}+1}}{\sqrt{x-1}+1}\)
\(=\dfrac{\left|\sqrt{x-1}-1\right|}{\sqrt{x-1}-1}+\dfrac{\sqrt{x-1}+1}{\sqrt{x-1}+1}\)
\(=\dfrac{\sqrt{x-1}-1}{\sqrt{x-1}-1}+1=1+1=2\)
c:
\(=\dfrac{a+b}{b^2}\cdot\dfrac{ab^2}{a+b}=a\)
d: Sửa đề: \(A=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(1+\sqrt{a}+a+\sqrt{a}\right)\cdot\left(\dfrac{1}{1+\sqrt{a}}\right)^2\)
\(=\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)^2}=1\)
e:
\(A=\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(\sqrt{y}-1\right)^2}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y}-1}{\left(x-1\right)^2}=\dfrac{1}{x-1}\)
f:
\(A=\sqrt{\dfrac{m}{\left(1-x\right)^2}\cdot\dfrac{4m\left(1-2x+x^2\right)}{81}}\)
\(=\sqrt{\dfrac{m}{\left(x-1\right)^2}\cdot\dfrac{4m\left(x-1\right)^2}{81}}\)
\(=\sqrt{\dfrac{4m^2}{81}}=\dfrac{2m}{9}\)