Giải phương trình
a) \(x^4+\sqrt{x^2+1995}=1995\)
b) \(\sqrt{4x+1}-\sqrt{3x+2}=\dfrac{x+3}{5}\)
c) \(\sqrt{x-4}-\sqrt{x+11}=-3\)
Giải phương trình
a) \(x^4+\sqrt{x^2+1995}=1995\)
b) \(\sqrt{4x+1}-\sqrt{3x+2}=\dfrac{x+3}{5}\)
c) \(\sqrt{x-4}-\sqrt{x+11}=-3\)
Giải phương trình
a) \(x^4+\sqrt{x^2+1995}=1995\)
b) \(\sqrt{4x+1}-\sqrt{3x+2}=\dfrac{x+3}{5}\)
c) \(\sqrt{x-4}-\sqrt{x+11}=-3\)
a) đặc : \(x^2=t\left(t\ge0\right)\)
\(\Rightarrow pt\Leftrightarrow t^2+\sqrt{t+1995}=1995\)
\(\Leftrightarrow\sqrt{t+1995}=1995-t^2\)
\(\Leftrightarrow t^4-3990t^2-t+1995.1994\)
\(\Leftrightarrow t^4+t^3-1994t^2-t^3-t^2+1994t-1995t^2-1995t+1995.1994=0\)
\(\Leftrightarrow t^2\left(t^2+t-1994\right)-t\left(t+t-1994\right)-1995\left(t^2+t-1994\right)=0\)
\(\Leftrightarrow\left(t^2-t-1995\right)\left(t^2+t-1994\right)=0\)
===> ...
câu b và c tương tự mấy câu bên kia nha
Mysterious Person Nguyễn Thanh Hằng Arakawa Whiter giup mk nha!!
Giải phương trình
a) \(x^4+\sqrt{x^2+1995}=1995\)
b) \(\sqrt{4x+1}-\sqrt{3x+2}=\dfrac{x+3}{5}\)
c) \(\sqrt{x-4}-\sqrt{x+11}=-3\)
a) đặc : \(x^2=t\left(t\ge0\right)\)
\(\Rightarrow pt\Leftrightarrow t^2+\sqrt{t+1995}=1995\)
\(\Leftrightarrow\sqrt{t+1995}=1995-t^2\)
\(\Leftrightarrow t^4-3990t^2-t+1995.1994\)
\(\Leftrightarrow t^4+t^3-1994t^2-t^3-t^2+1994t-1995t^2-1995t+1995.1994=0\)
\(\Leftrightarrow t^2\left(t^2+t-1994\right)-t\left(t+t-1994\right)-1995\left(t^2+t-1994\right)=0\)
\(\Leftrightarrow\left(t^2-t-1995\right)\left(t^2+t-1994\right)=0\)
===> ...
câu b và c tương tự mấy câu bên kia nha
Giải phương trình
a) \(x^4+\sqrt{x^2+1995}=1995\)
b) \(\sqrt{4x+1}-\sqrt{3x+2}=\dfrac{x+3}{5}\)
c) \(\sqrt{x-4}-\sqrt{x+11}=-3\)
b: \(\Leftrightarrow4x+1+3x+2-2\sqrt{\left(4x+1\right)\left(3x+2\right)}=\dfrac{1}{25}\left(x^2+6x+9\right)\)
=>1/25(x^2+6x+9)=7x+3-2căn(4x+1)(3x+2)
\(\Leftrightarrow x^2\cdot\dfrac{1}{25}+\dfrac{6}{25}x+\dfrac{9}{25}-7x-3=-2\sqrt{\left(4x+1\right)\left(3x+2\right)}\)
\(\Leftrightarrow\sqrt{4\left(4x+1\right)\left(3x+2\right)}=-\dfrac{1}{25}x^2-\dfrac{6}{25}x-\dfrac{9}{25}+7x+3=-\dfrac{1}{25}x^2+\dfrac{169}{25}x+\dfrac{66}{25}\)
\(\Leftrightarrow4\left(4x+1\right)\left(3x+2\right)=\left(\dfrac{1}{25}x^2-\dfrac{169}{25}x-\dfrac{66}{25}\right)^2\)
\(\Leftrightarrow x\in\left\{-0.13\right\}\)
c: \(\Leftrightarrow\sqrt{x+11}-\sqrt{x-4}=3\)
=>x+11+x-4-2căn (x+11)(x-4)=9
=>2căn(x+11)(x-4)=2x+7-9=2x-2
=>căn (x+11)(x-4)=x-1
=>x^2-2x+1=x^2-4x+11x-44
=>-2x+1=7x-44
=>-9x=-45
=>x=5
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. \(\sqrt{x^2+2x}=\sqrt{3x^2+4x+1}-\sqrt{3x^2+4x+1}\)
Giải phương trình
\(a.\dfrac{3}{4}\sqrt{4x}-\sqrt{4x}+5=\dfrac{1}{4}\sqrt{4x}\)
\(b.\sqrt{3-x}-\sqrt{27-9x}+1,25.\sqrt{48-16x}=6\)
\(c.\dfrac{5\sqrt{x}-2}{8\sqrt{x}+2,5}=\dfrac{2}{7}\)
\(d.\sqrt{9x^2+12x+4}=4\)
d. \(\sqrt{9x^2+12x+4}=4\)
<=> \(\sqrt{\left(3x+2\right)^2}=4\)
<=> \(|3x+2|=4\)
<=> \(\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)
c: Ta có: \(\dfrac{5\sqrt{x}-2}{8\sqrt{x}+2.5}=\dfrac{2}{7}\)
\(\Leftrightarrow35\sqrt{x}-14=16\sqrt{x}+5\)
\(\Leftrightarrow x=1\)
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 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 các phương trình sau:
a) \(\sqrt{x^2-4+4}=2-x\)
b) \(\sqrt{4x-8}-\dfrac{1}{5}\sqrt{25x-50}=3\sqrt{x-2}-1\)
c) \(\sqrt{x-1}+\sqrt{9x-9}-\sqrt{4x-4}=4\)
d) \(\dfrac{1}{2}\sqrt{x-2}-4\sqrt{\dfrac{4x-8}{9}}+\sqrt{9x-18}-5=0\)
e)\(\sqrt{49-28x+4x^2}-5=0\)
f) \(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\)
g) x2 - 4x - 2\(\sqrt{2x-5}+5=0\)
h)\(\sqrt{3x-2}=\sqrt{x+1}\)
i) x + y + z + 8 = \(2\sqrt{x-1}+4\sqrt{y-2}+6\sqrt{z-3}\)
k) \(\sqrt{x^2-3x}-\sqrt{x-3}=0\)
l)\(\sqrt{x^2-4}+\sqrt{x-2}=0\)
m) \(4\sqrt{x+1}=x^2-5x+14\)
n) \(\sqrt{x^2-6x+9}-\sqrt{4x^2+4x+1}=0\)
c: Ta có: \(\sqrt{x-1}+\sqrt{9x-9}-\sqrt{4x-4}=4\)
\(\Leftrightarrow2\sqrt{x-1}=4\)
\(\Leftrightarrow x-1=4\)
hay x=5
e: Ta có: \(\sqrt{4x^2-28x+49}-5=0\)
\(\Leftrightarrow\left|2x-7\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-7=5\\2x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=1\end{matrix}\right.\)
a. ĐKXĐ: $x\in\mathbb{R}$
PT $\Leftrightarrow \sqrt{(x-2)^2}=2-x$
$\Leftrightarrow |x-2|=2-x$
$\Leftrightarrow 2-x\geq 0$
$\Leftrightarrow x\leq 2$
b. ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow \sqrt{4}.\sqrt{x-2}-\frac{1}{5}\sqrt{25}.\sqrt{x-2}=3\sqrt{x-2}-1$
$\Leftrightarrow 2\sqrt{x-2}-\sqrt{x-2}=3\sqrt{x-2}-1$
$\Leftrightarrow 1=2\sqrt{x-2}$
$\Leftrightarrow \frac{1}{2}=\sqrt{x-2}$
$\Leftrightarrow \frac{1}{4}=x-2$
$\Leftrightarrow x=\frac{9}{4}$ (tm)
c. ĐKXĐ: $x\geq 1$
PT $\Leftrightarrow \sqrt{x-1}+\sqrt{9}.\sqrt{x-1}-\sqrt{4}.\sqrt{x-1}=4$
$\Leftrightarrow \sqrt{x-1}+3\sqrt{x-1}-2\sqrt{x-1}=4$
$\Leftrightarrow 2\sqrt{x-1}=4$
$\Leftrightarrow \sqrt{x-1}=2$
$\Leftrightarrow x-1=4$
$\Leftrightarrow x=5$ (tm)
d. ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow \frac{1}{2}\sqrt{x-2}-4\sqrt{\frac{4}{9}}\sqrt{x-2}+\sqrt{9}.\sqrt{x-2}-5=0$
$\Leftrightarrow \frac{1}{2}\sqrt{x-2}-\frac{8}{3}\sqrt{x-2}+3\sqrt{x-2}-5=0$
$\Leftrightarrow \frac{5}{6}\sqrt{x-2}-5=0$
$\Leftrightarrow \sqrt{x-2}=6$
$\Leftrightarrow x-2=36$
$\Leftrightarrow x=38$ (tm)