1) \(2\sqrt{x+3}=x-1+4\sqrt{2x-1}\)
2) \(\sqrt[4]{x-1}+\sqrt[4]{5-x}=2\)
3) \(\sqrt[3]{1-2x}+\sqrt{x+3}=1\)
4) \(\dfrac{\sqrt{x}}{1+\sqrt{1-x}}=x^2-2x+2\)
1) \(x+\sqrt{1-x^2}< x\sqrt{1-x^2}\)
2)\(\dfrac{1}{\sqrt{2x^2+3x-3}}>\dfrac{1}{2x-1}\)
3)\(5\sqrt{x}+\dfrac{5}{2\sqrt{x}}< 2x+\dfrac{1}{2x}+4\)
giúp mình ạ
Giải các phương trình sau:
1) \(\sqrt{3x^2+5x+8}-\sqrt{3x^2+5x+1}=1\)
2) \(x^2-2x-12+4\sqrt{\left(4-x\right)\left(2+x\right)}=0\)
3) \(3\sqrt{x}+\dfrac{3}{2\sqrt{x}}=2x+\dfrac{1}{2x}-7\)
4) \(\sqrt{x}-\dfrac{4}{\sqrt{x+2}}+\sqrt{x+2}=0\)
5)\(\left(x-7\right)\sqrt{\dfrac{x+3}{x-7}}=x+4\)
6) \(2\sqrt{x-4}+\sqrt{x-1}=\sqrt{2x-3}+\sqrt{4x-16}\)
7) \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\dfrac{x+3}{2}\)
Giúp mình với ajk, mink đang cần gấp
Giải phương trình:
1. \(5x^2+2x+10=7\sqrt{x^4+4}\)
2. \(\dfrac{4}{x}+\sqrt{x-\dfrac{1}{x}}=x+\sqrt{2x-\dfrac{5}{x}}\)
3. \(\sqrt{x^2+2x}=\sqrt{3x^2+4x+1}-\sqrt{3x^2+4x+1}\)
\(\lim\limits_{x\rightarrow+\infty}\dfrac{x\sqrt{x^2+1}+2x+1}{\sqrt[3]{2x^3+x+1}+x}\)
\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{2x^2-x+1}-\sqrt[3]{2x+3}}{3x^2-2}\)
\(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{4x^2+x}+\sqrt[3]{8x^3+x-1}}{\sqrt[4]{x^4+3}}\)
a/ \(\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{x}{x}\sqrt{x^2+1}+\dfrac{2x}{x}+\dfrac{1}{x}}{\dfrac{x}{x}\sqrt[3]{\dfrac{2x^3}{x^3}+\dfrac{x}{x^3}+\dfrac{1}{x^3}}+\dfrac{x}{x}}=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{x^2+1}+2}{\sqrt[3]{2}+1}=+\infty\)
b/ \(=\lim\limits_{x\rightarrow1}\dfrac{\sqrt{2.1^2-1+1}-\sqrt[3]{2.1+3}}{3.1^2-2}=...\)
c/ \(\lim\limits_{x\rightarrow+\infty}\dfrac{x\sqrt{\dfrac{4x^2}{x^2}+\dfrac{x}{x^2}}+x\sqrt[3]{\dfrac{8x^3}{x^3}+\dfrac{x}{x^3}-\dfrac{1}{x^3}}}{x\sqrt[4]{\dfrac{x^4}{x^4}+\dfrac{3}{x^4}}}=\dfrac{2+2}{1}=4\)
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
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}`
Giải các phương trình sau:
1. \(\sqrt{x^2-\dfrac{1}{4}+\sqrt{x^2+x+\dfrac{1}{4}}}=\dfrac{1}{2}\left(2x^3+x^2+2x+1\right)\)
2. \(x^2+4x+7=\left(x+4\right)\sqrt{x^2+7}\)
3. \(\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+\sqrt{x^4-1}\)
4. \(\sqrt{x^2-3x+2}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{x^2+2x-3}\)
5. \(x=\left(\sqrt{x}+2\right)\left(1-\sqrt{1-\sqrt{x}}\right)\)
6. \(2\sqrt[3]{2x-1}=x^3+1\)
7. \(\sqrt{x-\dfrac{1}{x}}+\sqrt{1-\dfrac{1}{x}}=x\)
Tìm x, biết:
a) \(\sqrt{x^2-2x+1}=2\)
b)\(\sqrt{x^2-1}=x\)
c) \(\sqrt{4x-20}+3\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
d) \(x-5\sqrt{x-2}=-2\)
e) \(2x-3\sqrt{2x-1}-5=0\)
`a)sqrt{x^2-2x+1}=2`
`<=>sqrt{(x-1)^2}=2`
`<=>|x-1|=2`
`**x-1=2<=>x=3`
`**x-1=-1<=>x=-1`.
Vậy `S={3,-1}`
`b)sqrt{x^2-1}=x`
Điều kiện:\(\begin{cases}x^2-1 \ge 0\\x \ge 0\\\end{cases}\)
`<=>` \(\begin{cases}x^2 \ge 1\\x \ge 0\\\end{cases}\)
`<=>x>=1`
`pt<=>x^2-1=x^2`
`<=>-1=0` vô lý
Vậy pt vô nghiệm
`c)sqrt{4x-20}+3sqrt{(x-5)/9}-1/3sqrt{9x-45}=4(x>=5)`
`pt<=>sqrt{4(x-5)}+sqrt{9*(x-5)/9}-sqrt{(9x-45)*1/9}=4`
`<=>2sqrt{x-5}+sqrt{x-5}-sqrt{x-5}=4`
`<=>2sqrt{x-5}=4`
`<=>sqrt{x-5}=2`
`<=>x-5=4`
`<=>x=9(tmđk)`
Vậy `S={9}.`
`d)x-5sqrt{x-2}=-2(x>=2)`
`<=>x-2-5sqrt{x-2}+4=0`
Đặt `a=sqrt{x-2}`
`pt<=>a^2-5a+4=0`
`<=>a_1=1,a_2=4`
`<=>sqrt{x-2}=1,sqrt{x-2}=4`
`<=>x_1=3,x_2=18`,
`e)2x-3sqrt{2x-1}-5=0`
`<=>2x-1-3sqrt{2x-1}-4=0`
Đặt `a=sqrt{2x-1}(a>=0)`
`pt<=>a^2-3a-4=0`
`a-b+c=0`
`<=>a_1=-1(l),a_2=4(tm)`
`<=>sqrt{2x-1}=4`
`<=>2x-1=16`
`<=>x=17/2(tm)`
Vậy `S={17/2}`
d.
ĐKXĐ: $x\geq 2$. Đặt $\sqrt{x-2}=a(a\geq 0)$ thì pt trở thành:
$a^2+2-5a=-2$
$\Leftrightarrow a^2-5a+4=0$
$\Leftrightarrow (a-1)(a-4)=0$
$\Rightarrow a=1$ hoặc $a=4$
$\Leftrightarrow \sqrt{x-2}=1$ hoặc $\sqrt{x-2}=4$
$\Leftrightarrow x=3$ hoặc $x=18$ (đều thỏa mãn)
e. ĐKXĐ: $x\geq \frac{1}{2}$
Đặt $\sqrt{2x-1}=a(a\geq 0)$ thì pt trở thành:
$a^2+1-3a-5=0$
$\Leftrightarrow a^2-3a-4=0$
$\Leftrightarrow (a+1)(a-4)=0$
Vì $a\geq 0$ nên $a=4$
$\Leftrightarrow \sqrt{2x-1}=4$
$\Leftrightarrow x=\frac{17}{2}$
a.
$\sqrt{x^2-2x+1}=2$
$\Leftrightarrow \sqrt{(x-1)^2}=2$
$\Leftrightarrow |x-1|=2$
$\Rightarrow x-1=\pm 2$
$\Leftrightarrow x=3$ hoặc $x=-1$ (đều thỏa mãn)
b. ĐKXĐ: $x\geq 1$ hoặc $x\leq -1$
PT \(\Rightarrow \left\{\begin{matrix} x\geq 0\\ x^2-1=x^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 0\\ 1=0\end{matrix}\right.\) (vô lý)
Vậy pt vô nghiệm
c. ĐKXĐ: $x\geq 5$
PT $\Leftrightarrow \sqrt{4(x-5)}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9(x-5)}=4$
$\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4$
$\Leftrightarrow 2\sqrt{x-5}=4$
$\Leftrightarrow \sqrt{x-5}=2$
$\Leftrightarrow x=2^2+5=9$ (thỏa mãn)
1.\(\sqrt{-4x^2+25}=x\)
2.\(\sqrt{3x^2-4x+3}=1-2x\)
3. \(\sqrt{4\left(1-x\right)^2}-\sqrt{3}=0\)
4.\(\dfrac{3\sqrt{x+5}}{\sqrt{ }x-1}< 0\)
5. \(\dfrac{3\sqrt{x-5}}{\sqrt{x+1}}\ge0\)
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/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{\dfrac{4x^2}{x^2}+\dfrac{x}{x^2}}+\sqrt[3]{\dfrac{8x^3}{x^3}+\dfrac{x}{x^3}-\dfrac{1}{x^3}}}{\sqrt[4]{\dfrac{x^4}{x^4}+\dfrac{3}{x^4}}}=\dfrac{2+2}{1}=4\)