giải pt :
a,\(2x^2-11x+21=3\sqrt[3]{4x-4}\)
b,\(\dfrac{\sqrt{x-3}}{\sqrt{2x-1}-1}=\dfrac{1}{\sqrt{x+3}-\sqrt{x-3}}\)
c,\(\left(\sqrt{x^2+x+1}+\sqrt{4x^2+x+1}\right)\left(\sqrt{5x^2+1}-\sqrt{2x^2+1}\right)=3x^2\)
giải phương trình :
a, \(\dfrac{4x-1}{\sqrt{4x-3}}+\dfrac{11-2x}{\sqrt{5-x}}=\dfrac{15}{2}\)
b, \(\left(\sqrt{5x-1}+\sqrt{x-1}\right)\left(3x-1-\sqrt{5x^2-6x+1}\right)=4x\)
giải pt :
a, \(\dfrac{\sqrt{x-3}}{\sqrt{2x-1}-1}=\dfrac{1}{\sqrt{x+3}-\sqrt{x-3}}\)
b, \(\left(\sqrt{x^2+x+1}+\sqrt{4x^2+x+1}\right)\left(\sqrt{5x^2+1}-\sqrt{2x^2+1}\right)=3x^2\)
Giải phương trình sau:
a) \(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)
b) \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
c) \(2x-x^2+\sqrt{6x^2-12x+7}=0\)
d) \(\left(x+1\right)\left(x+4\right)-3\sqrt{x^2+5x+2}=6\)
a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow x+5=4\)
hay x=-1
b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\Leftrightarrow\sqrt{x-1}=17\)
\(\Leftrightarrow x-1=289\)
hay x=290
Tìm giới hạn
1) \(\xrightarrow[x->3]{lim}\dfrac{x^2-5x+6}{\sqrt{2x+3}-\sqrt{4x-3}}\)
2) \(\xrightarrow[x->1]{lim}\dfrac{\sqrt{x^2+2}-\sqrt{4x-1}}{x-1}\)
3) \(\xrightarrow[x->-1]{lim}\dfrac{x-2}{x\left|x+1\right|}\)
4) \(\xrightarrow[x->a]{lim}\dfrac{x^n-a^n}{x-a}\)
5) \(\xrightarrow[x->1]{lim}(\dfrac{n}{1-x^n}-\dfrac{1}{1-x})\)
6) \(\xrightarrow[x->1]{lim}\dfrac{x^n-nx+n-1}{\left(x-1\right)^2}\)
a) lim \(\dfrac{x\sqrt{x^2+1}-2x+1}{^3\sqrt{2x^3-2}+1}\)
x-> -∞
b) lim \(\dfrac{\left(2x+1\right)^3\left(x+2\right)^4}{\left(3-2x\right)^7}\)
x-> -∞
c) lim \(\dfrac{\sqrt{4x^2+x}+^3\sqrt{8x^3+x-1}}{^4\sqrt{x^4+3}}\)
x-> +∞
a/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{x\sqrt{x^2+1}}{x}-\dfrac{2x}{x}+\dfrac{1}{x}}{\sqrt[3]{\dfrac{2x^3}{x^3}-\dfrac{2x}{x^3}}+\dfrac{1}{x}}=0\)
b/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{8x^7}{x^7}}{\dfrac{\left(-2x^7\right)}{x^7}}=-\dfrac{8}{2^7}\)
c/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{\dfrac{4x^2}{x^2}+\dfrac{x}{x^2}}+\sqrt[3]{\dfrac{8x^3}{x^3}+\dfrac{x}{x^3}-\dfrac{1}{x^3}}}{\sqrt[4]{\dfrac{x^4}{x^4}+\dfrac{3}{x^4}}}=\dfrac{2+2}{1}=4\)
Tìm điều kiện có nghĩa:
1) \(-\dfrac{1}{\sqrt{a+2}}\)
2) \(\sqrt{\dfrac{3}{\left(x-2\right)^2}}\)
3) \(\sqrt{\dfrac{-3}{a^2-4a+4}}\)
4) \(\sqrt{\dfrac{2}{x^2+2x+2}}\)
5) \(\sqrt{\dfrac{-3}{x^2-4x+5}}\)
6) \(\sqrt{\dfrac{-4}{x^2-1}}\)
7) \(\sqrt{\dfrac{x+1}{x-2}}\)
8) \(\sqrt{\dfrac{x-2}{x+3}}\)
1: ĐKXĐ: \(a>-2\)
2: ĐKXĐ: \(x\ne2\)
3: ĐKXĐ: \(a\in\varnothing\)
1)
\(-\dfrac{1}{\sqrt{a+2}}\) có nghĩa khi \(\sqrt{a+2}>0\)
=>a+2>0
a>-2
2)
\(\sqrt{\dfrac{3}{\left(x-2\right)^2}}=\dfrac{\sqrt{3}}{\sqrt{\left(x-2\right)^2}}\)
mà \(\left(x-2\right)^2>0=>\sqrt{\left(x-2\right)^2}>0vớimọix\)
3)
\(\sqrt{\dfrac{-3}{a^2-4a+4}}=\sqrt{\dfrac{-3}{\left(a-2\right)^2}}cónghĩakhi\left(a-2\right)^2< 0mà\left(a-2\right)^2>0=>biểuthứckocónghĩavớimọia\)
a) \(-\dfrac{1}{\sqrt{a+2}}\Rightarrow\sqrt{a+2}>0\Leftrightarrow a>-2\)
b) \(\sqrt{\dfrac{3}{\left(x-2\right)^2}}\Rightarrow x-2\ne0\Leftrightarrow x\ne2\)
c) \(\sqrt{\dfrac{-3}{a^2-4a+4}}\Rightarrow a^2-4a+4< 0\Leftrightarrow a^2-4a< -4\)
d) \(\sqrt{\dfrac{2}{x^2+2x+2}}\Rightarrow x^2+2x+2>0\Leftrightarrow x^2+2x>-2\)
e) \(\sqrt{\dfrac{-3}{x^2-4x+5}}\Rightarrow x^2-4x+5< 0\Leftrightarrow x^2-4x< -5\)
f) \(\sqrt{\dfrac{-4}{x^2-1}}\Rightarrow x^2-1< 0\Rightarrow x^2< 1\Rightarrow x< 1\)
2 câu cuối do lỗi nên mk ko gõ cth được
Giải :
a) \(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{2x+3}}=\sqrt{3}\left(\dfrac{1}{\sqrt{4x}-3}+\dfrac{1}{\sqrt{5x}-6}\right)\)
b) \(x+\dfrac{4}{x+2}=3\)
b)\(x+\dfrac{4}{x+2}=3\) (ĐKXĐ: \(x\ne-2\))
\(\Leftrightarrow x+\dfrac{4}{x+2}-3=0\\ \Leftrightarrow\dfrac{x\left(x+2\right)+4-3\left(x+2\right)}{x+2}=0\\ \Leftrightarrow\dfrac{x^2+2x+4-3x-6}{x+2}=0\\ \Leftrightarrow\dfrac{x^2-x-2}{x+2}=0\\ \Leftrightarrow x^2-x-2=0\\ \Leftrightarrow x=\dfrac{-\left(-1\right)\pm\sqrt{\left(-1\right)^2-4.1.\left(-2\right)}}{2.1}\\ \Leftrightarrow x=\dfrac{1\pm\sqrt{1+8}}{2}\\ \Leftrightarrow x=\dfrac{1\pm\sqrt{9}}{2}\\ \Leftrightarrow x=\dfrac{1\pm3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+3}{2}\\x=\dfrac{1-3}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(TM\right)\\x=-1\left(TM\right)\end{matrix}\right.\)
Vậy...
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{4x+1}-\sqrt[3]{2x+1}}{x}\)
\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{4x+5}-3}{\sqrt[3]{5x+3}-2}\)
\(\lim\limits_{x\rightarrow-1}\dfrac{\sqrt[4]{2x+3}+\sqrt[3]{2+3x}}{\sqrt{x+2}-1}\)
\(a=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{4x+1}-1+1-\sqrt[3]{2x+1}}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{4x}{\sqrt[]{4x+1}+1}+\dfrac{-2x}{1+\sqrt[3]{2x+1}+\sqrt[3]{\left(2x+1\right)^2}}}{x}\)
\(=\lim\limits_{x\rightarrow0}\left(\dfrac{4}{\sqrt[]{4x+1}+1}+\dfrac{-2}{1+\sqrt[3]{2x+1}+\sqrt[3]{\left(2x+1\right)^2}}\right)=...\)
\(b=\lim\limits_{x\rightarrow1}\dfrac{4\left(x-1\right)\left(\sqrt[3]{\left(5x+3\right)^2}+2\sqrt[3]{5x+3}+4\right)}{5\left(x-1\right)\left(\sqrt[]{4x+5}+3\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{4\left(\sqrt[3]{\left(5x+3\right)^2}+2\sqrt[3]{5x+3}+4\right)}{5\left(\sqrt[]{4x+5}+3\right)}=...\)
\(c=\lim\limits_{x\rightarrow-1}\dfrac{\left(2x+3\right)^{\dfrac{1}{4}}+\left(2+3x\right)^{\dfrac{1}{3}}}{\left(x+2\right)^{\dfrac{1}{2}}-1}\)
\(=\lim\limits_{x\rightarrow-1}\dfrac{\dfrac{1}{2}\left(2x+3\right)^{-\dfrac{3}{4}}+\left(2+3x\right)^{-\dfrac{2}{3}}}{\dfrac{1}{2}\left(x+2\right)^{-\dfrac{1}{2}}}=3\)
