Tìm \(x\) thỏa mãn điều kiện
a) \(\sqrt{\dfrac{2x-3}{x-1}}=2\)
b) \(\dfrac{\sqrt{2x-3}}{\sqrt{x-1}}=2\)
c) \(\sqrt{\dfrac{4x+3}{x+1}}=3\)
d) \(\dfrac{\sqrt{4x+3}}{\sqrt{x+1}}=3\)
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
Tìm x để mỗi căn thức sau có nghĩa:
a. \(\sqrt{3-2x}\) b. \(\sqrt{x+1}+\sqrt{3-x}\) c. \(\dfrac{\sqrt{4x-2}}{x^2-4x+3}\) d. \(\dfrac{\sqrt{4x^2-2x+1}}{\sqrt{3-5x}}\)
ĐKXĐ: \(3-2x\ge0\Leftrightarrow x\le\dfrac{3}{2}\)
b) ĐKXĐ: \(-1\le x\le3\)
c) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x\ne1\\x\ne3\end{matrix}\right.\).
d) ĐKXĐ: \(x< \dfrac{3}{5}\).
1.
a. Tìm điều kiện để căn thức bậc hai có nghĩa \(\sqrt{\dfrac{x^2}{2x-1}}\)
b. \(\dfrac{\sqrt[3]{625}}{\sqrt[3]{5}}-\sqrt[3]{-216}.\sqrt[3]{\dfrac{1}{27}}\)
* Giải phương trình
a. \(\sqrt{\left(x+1\right)^2}=3\)
b. \(3\sqrt{4x+4}-\sqrt{9x+9}-8\sqrt{\dfrac{x+1}{16}}=5\)
Tìm điều kiện có nghĩa:
1) \(\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
2) \(\sqrt{\dfrac{2}{x^2+2x+2}}\)
3) \(\sqrt{\dfrac{-3}{x^2-4x+5}}\)
1) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\y\ge0\\\sqrt{x}+\sqrt{y}\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x>0\\y>0\end{matrix}\right.\)
2) ĐKXĐ: \(x^2+2x+2>0\Leftrightarrow\left(x^2+2x+1\right)+1>0\Leftrightarrow\left(x+1\right)^2+1>0\left(đúng\forall x\right)\)
3) ĐKXĐ: \(x^2-4x+5< 0\Leftrightarrow\left(x^2-4x+4\right)+1< 0\Leftrightarrow\left(x-2\right)^2+1< 0\left(VLý.do.\left(x-2\right)^2+1\ge1>0\right)\)
Vậy biểu thức không xác định với mọi x
Đkien
a) \(\left\{{}\begin{matrix}x\ge0;y\ge0\\\sqrt[]{x}+\sqrt{y}\ne0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge0,y>0\\x>0,y\ge0\end{matrix}\right.\)
b) \(\dfrac{2}{x^2+2x+2}\ge0\Leftrightarrow x^2+2x+2>0\)
\(\Leftrightarrow x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}>0\forall x\)
=> PT luôn xác định
c) \(-\dfrac{3}{x^2-4x+5}\ge0\Leftrightarrow x^2-4x+5< 0\)
\(\)=> vô nghiệm
Vậy căn thức k xác định
Tìm điều kiện xác định và rút gọn các biểu thức sau :
a/ \(A=\left(\dfrac{\sqrt{3}}{x^2+x\sqrt{3}+3}+\dfrac{3}{x^3-\sqrt{27}}\right).\left(\dfrac{x}{\sqrt{3}}+\dfrac{\sqrt{3}}{x}+1\right)\)
b/ \(B=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+x+1\)
c/ \(C=\left(\dfrac{2+\sqrt{x}}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right).\dfrac{x\sqrt{x}+x-\sqrt{x}-1}{\sqrt{x}}\)
d/ \(\left[\dfrac{1}{x-1}+\dfrac{x^2+1-2x}{\left(x-1\right)^2+3x}-\dfrac{1+4x-2x^2}{x^3-1}\right]:\dfrac{2}{x^2+1}\)
a: \(A=\left(\dfrac{\sqrt{3}\left(x-\sqrt{3}\right)+3}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}\right)\cdot\dfrac{x^2+3+x\sqrt{3}}{x\sqrt{3}}\)
\(=\dfrac{x\sqrt{3}}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}\cdot\dfrac{x^2+x\sqrt{3}+3}{x\sqrt{3}}\)
\(=\dfrac{1}{x-\sqrt{3}}\)
b: \(B=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\)
\(=x-\sqrt{x}-x-\sqrt{x}+x+1\)
\(=x-2\sqrt{x}+1\)
c: \(C=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{x\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}-2-\left(x-\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}}=2\)
Tìm điều kiện để các biểu thức sau xác định
a)\(\sqrt{x+1}-\dfrac{1}{2}\)
b)\(2.\sqrt{1-2x}-\dfrac{\sqrt{3}-1}{4}\)
c)\(\sqrt{x+1}-\sqrt{x-2}\)
d)\(\sqrt{2-3x}-\sqrt{1-2x}\)
e)\(2.\sqrt{\sqrt{3}-2x}+\dfrac{1}{x-1}\)
f)\(\dfrac{1}{2}.\sqrt{x-\dfrac{\sqrt{3}}{2}}-\dfrac{1}{\sqrt{x}-1}\)
g)\(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+2}\)
h)\(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x^2+2}}\)
a, \(x+1\ge0\Leftrightarrow x\ge-1\)
b, \(1-2x\ge0\Leftrightarrow x\le\dfrac{1}{2}\)
c, \(\left\{{}\begin{matrix}x+1\ge0\\x-2\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\ge2\end{matrix}\right.\Leftrightarrow x\ge2\)
d, \(\left\{{}\begin{matrix}2-3x\ge0\\1-2x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{2}{3}\\x\le\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x\le\dfrac{1}{2}\)
e, \(\left\{{}\begin{matrix}\sqrt{3}-2x\ge0\\x-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{\sqrt{3}}{2}\\x\ne1\end{matrix}\right.\Leftrightarrow x\le\dfrac{\sqrt{3}}{2}\)
f, \(\left\{{}\begin{matrix}x-\dfrac{\sqrt{3}}{2}\ge0\\x\ge0\\\sqrt{x}-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{\sqrt{3}}{2}\\x\ge0\\x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{\sqrt{3}}{2}\\x\ne1\end{matrix}\right.\)
g, \(\left\{{}\begin{matrix}\sqrt{x}-1\ne0\\\sqrt{x}+2\ne0\\x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
giải pt :
a, (x+5)(2-x)=3\(\sqrt{x^2+3x}\)
b, \(\sqrt[3]{\dfrac{2x}{x+1}}+\sqrt[3]{\dfrac{1}{2}+\dfrac{1}{2x}}=2\)
c,\(\sqrt[5]{\dfrac{16x}{x-1}}+\sqrt[5]{\dfrac{x-1}{16x}}=\dfrac{5}{2}\)
d, \(\sqrt{5x^2+10x+1}=7-2x-x^2\)
e, \(\sqrt{2x^2+4x+1}=1-2x-x^2\)
Giải phương trình:
a) \(\sqrt{x}+\sqrt{2-x}=\dfrac{3x^2-2x+3}{x^2+1}\)
b) \(x^3-11x^2+36x-18=4\sqrt[4]{27x-54}\)
c) \(16x^4+5=6\sqrt[3]{4x^3+x}\)
d) \(\dfrac{1}{\sqrt{2x-1}}+\dfrac{1}{\sqrt[4]{4x-3}}=\dfrac{2}{x}\)
b, \(đk:x\ge2\)
Xét x=2 thay vào pt thấy không thỏa mãn => x>2 hay 27x-54>0
\(x^3-11x+36x-18=4\sqrt[4]{27x-54}\)
\(\Leftrightarrow27x^3-297x^2+972x-486=4\sqrt[4]{\left(27x-54\right).81.81.81}\le189+27x\) (cosi với 4 số dương, dấu = xảy ra khi x=5)
\(\Leftrightarrow x^3-11x^2+35x-25\le0\)
\(\Leftrightarrow\left(x-1\right)\left(x-5\right)^2\le0\) (*)
Có \(\left\{{}\begin{matrix}x>2\\\left(x-5\right)^2\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-1>0\\\left(x-5\right)^2\ge0\end{matrix}\right.\)\(\Rightarrow\left(x-1\right)\left(x-5\right)^2\ge0\) (2*)
Từ (*) và (2*) ,dấu = xra khi x=5 (thỏa mãn)
Vây pt có nghiệm duy nhất x=5
c,Có \(6\sqrt[3]{4x^3+x}=16x^4+5>0\)
\(\Leftrightarrow4x^3+x>0\)
Có: \(16x^4+5=6\sqrt[3]{4x^3+x}\le2\left(4x^3+x+2\right)\) (theo cosi với 3 số dương,dấu = xảy ra khi \(x=\dfrac{1}{2}\))
\(\Leftrightarrow16x^4-8x^3-2x+1\le0\)
\(\Leftrightarrow\left(2x-1\right)^2\left(4x^2+2x+1\right)\le0\) (*)
(tương tự câu b) Dấu = xảy ra khi \(x=\dfrac{1}{2}\)(thỏa mãn)
Vậy....
d) Đk: \(x\ge\dfrac{3}{4}\)
Áp dụng bđt cosi:
\(\sqrt{2x-1}\le\dfrac{2x-1+1}{2}=x\)
\(\Rightarrow\dfrac{1}{\sqrt{2x-1}}\ge\dfrac{1}{x}\) (*)
\(\sqrt[4]{4x-3}\le\dfrac{4x-3+1+1+1}{4}=x\)
\(\dfrac{\Rightarrow1}{\sqrt[4]{4x-3}}\ge\dfrac{1}{x}\) (2*)
Từ (*) và (2*) \(\Rightarrow\dfrac{1}{\sqrt{2x-1}}+\dfrac{1}{\sqrt[4]{4x-3}}\ge\dfrac{2}{x}\)
Dấu = xảy ra khi x=1 (tm)
`a)\sqrtx+\sqrt{2-x}=(3x^2-2x+3)/(x^2+1)`
`đk:0<=x<=2`
`pt<=>sqrtx-1+\sqrt{2-x}-1=(3x^2-2x+3)/(x^2+1)-2`
`<=>(x-1)/(sqrtx+1)+(1-x)/(sqrt{2-x}+1)=(x^2-2x+1)/(x^2+1)`
`<=>(x-1)/(sqrtx+1)+(1-x)/(sqrt{2-x}+1)=(x-1)^2/(x^2+1)`
`<=>(x-1)((x-1)/(x^2+1)+1/(sqrt{2-x}+1)-1/(sqrtx+1))=0`
`<=>x-1=0<=>x=1`
Vậy `S={1}`
tìm điều kiện bài toán:
a) \(y=\dfrac{1}{x}-\dfrac{\sqrt{2x-1}}{x^2-3x+2}\)
b) \(y=\dfrac{1}{x^2-1}-\sqrt{7-2x}\)
c) \(y=\dfrac{2}{x}+\dfrac{3}{4-2x+x^2}\)
d) \(y=\sqrt{25-x^2}-2\sqrt{x}+3\)
Lời giải:
a.
\(\left\{\begin{matrix} x\neq 0\\ 2x-1\geq 0\\ x^2-3x+2=(x-1)(x-2)\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\neq 0\\ x\geq \frac{1}{2}\\ x\neq 1; x\neq 2\end{matrix}\right.\)
$\Leftrightarrow x\geq \frac{1}{2}; x\neq 1; x\neq 2$
b. \(\left\{\begin{matrix}
x^2-1=(x-1)(x+1)\neq 0\\
7-2x\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}
x\neq \pm 1\\
x\leq \frac{7}{2}\end{matrix}\right.\)
c.
\(\left\{\begin{matrix} x\neq 0\\ 4-2x+x^2\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\neq 0\\ (x-1)^2+3\neq 0\end{matrix}\right.\Leftrightarrow x\neq 0\)
d.
\(\left\{\begin{matrix} 25-x^2=(5-x)(5+x)\geq 0\\ x\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} -5\leq x\leq 5\\ x\geq 0\end{matrix}\right.\Leftrightarrow 0\leq x\leq 5\)
a) \(y=\dfrac{1}{x}-\dfrac{\sqrt[]{2x-1}}{x^2-3x+2}\)
Điều kiện \(\) \(2x-1\ge0;x\ne0;x^2-3x+2\ne0\)
\(\Leftrightarrow x\ge\dfrac{1}{2};x\ne0;\left(x-1\right)\left(x-2\right)\ne0\)
\(\Leftrightarrow x\ge\dfrac{1}{2};x\ne0;x\ne1;x\ne2\)
a) \(x\ge\dfrac{1}{2};x\ne1;x\ne2\)
b) \(x\le\dfrac{7}{2};x\ne\pm1\)
c) \(x\ne0\)
d) \(0\le x\le5\)