giải pt :
a, \(729x^4+8\sqrt{1-x^2}=36\)
b, \(3x^2-12x-5\sqrt{10+4x-x^2}+12=0\)
giải pt :
a,\(\left(6x-5\right)\sqrt{x+1}-\left(6x+2\right)\sqrt{x-1}+4\sqrt{x^2-1}=4x-3\)
b, \(\left(9x-2\right)\sqrt{3x-1}+\left(10-9x\right)\sqrt{3-3x}-4\sqrt{-9x^2+12x-3}=4\)
c, \(\left(13-4x\right)\sqrt{2x-3}+\left(4x-3\right)\sqrt{5-2x}=2+8\sqrt{-4x^2+16x-15}\)
giải pt:
a. \(\sqrt{x-2}+\sqrt{10-x}=x^2-12x+40\)
b. \(\sqrt{3x-5}+\sqrt{7-3x}=5x^2-20x+22\)
c. \(\sqrt{x^2-4x+4}+\sqrt{x^2-6x+9}=1\)
Giải các phương trình (giải chi tiết):
a) \(\sqrt{3x}-5\sqrt{12x}+7\sqrt{27x}=12\)
b) \(5\sqrt{9x+9}-2\sqrt{4x+4}+\sqrt{x+1}=36\)
`a)\sqrt{3x}-5\sqrt{12x}+7\sqrt{27x}=12` `ĐK: x >= 0`
`<=>\sqrt{3x}-10\sqrt{3x}+21\sqrt{3x}=12`
`<=>12\sqrt{3x}=12`
`<=>\sqrt{3x}=1`
`<=>3x=1<=>x=1/3` (t/m)
`b)5\sqrt{9x+9}-2\sqrt{4x+4}+\sqrt{x+1}=36` `ĐK: x >= -1`
`<=>15\sqrt{x+1}-4\sqrt{x+1}+\sqrt{x+1}=36`
`<=>12\sqrt{x+1}=36`
`<=>\sqrt{x+1}=3`
`<=>x+1=9`
`<=>x=8` (t/m)
1. Cho PT sau, tính x theo a, b (a,b > 0):
\(\sqrt{a+b\sqrt{x}}=2+\sqrt{a-b\sqrt{x}}\)
Áp dụng khi a = 2015, b = 2016 (làm tròn đến số thập phân thứ 5).
2. Giải PT:
\(729x^2+8\sqrt{1-x^2}=36\)
giải pt :
a,\(9x^2-6x-5=\sqrt{3x+5}\)
b, \(9x^2+12x-2=\sqrt{3x+8}\)
c, \(x^2-4x-3=\sqrt{x+5}\)
d,\(x^2-6x-2=\sqrt{x+8}\)
a.
ĐKXĐ: \(x\ge-\dfrac{5}{3}\)
\(9x^2-3x-\left(3x+5\right)-\sqrt{3x+5}=0\)
Đặt \(\sqrt{3x+5}=t\ge0\)
\(\Rightarrow9x^2-3x-t^2-t=0\)
\(\Delta=9+36\left(t^2+t\right)=\left(6t+3\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3+6t+3}{18}=\dfrac{t+1}{3}\\x=\dfrac{3-6t-3}{18}=-\dfrac{t}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}t=3x-1\\t=-3x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x+5}=3x-1\left(x\ge\dfrac{1}{3}\right)\\\sqrt{3x+5}=-3x\left(x\le0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+5=9x^2-6x+1\left(x\ge\dfrac{1}{3}\right)\\3x+5=9x^2\left(x\le0\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
c.
ĐKXĐ: \(x\ge-5\)
\(x^2-3x+2-x-5-\sqrt{x+5}=0\)
Đặt \(\sqrt{x+5}=t\ge0\)
\(\Rightarrow-t^2-t+x^2-3x+2=0\)
\(\Delta=1+4\left(x^2-3x+2\right)=\left(2x-3\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{1+2x-3}{-2}=1-x\\t=\dfrac{1-2x+3}{-2}=x-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+5}=1-x\left(x\le1\right)\\\sqrt{x+5}=x-2\left(x\ge2\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=x^2-2x+1\left(x\le1\right)\\x+5=x^2-4x+4\left(x\ge2\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
b.
ĐKXĐ: \(x\ge-\dfrac{8}{3}\)
\(\left(3x+2\right)^2-6-\sqrt{3x+8}=0\)
Đặt \(\sqrt{3x+8}=t\ge0\Rightarrow3x+2=t^2-6\)
\(\left(t^2-6\right)^2-6-t=0\)
\(\Leftrightarrow t^4-12t^2-t+30=0\)
\(\Leftrightarrow\left(t^2+t-5\right)\left(t^2-t-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=3\\t=\dfrac{\sqrt{21}-1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x+8}=3\\\sqrt{3x+8}=\dfrac{\sqrt{21}-1}{2}\end{matrix}\right.\)
\(\Leftrightarrow...\)
GIẢI CÁC PT SAU:
\(\sqrt{5x+10}=8-x\)
\(\sqrt{4x^2+x-12}=3x-5\)
\(\sqrt{x^2-2x+6}=2x-3\)
\(\sqrt{3x^2-2x+6}+3-2x=0\)
GIẢI CÁC PT SAU:
x2 - 6x + 9=\(4\sqrt{x^2-6x+6}\)
x2 - x + 8 - \(4\sqrt{x^2-x+4}=0\)
x2 + \(\sqrt{4x^2-12x+44}=3x+4\)
Giải PT \(729x^4+8\sqrt{1-x^2}=36\)
giải phương trình
a, \(\sqrt{1+x}-\sqrt{8-x}+\sqrt{\left(1+x\right)\left(8-x\right)}=3\)
b, \(\sqrt{3x^2+5x+8}-\sqrt{3x^2+5x+1}=1\)
c, \(2x^2+4x=\sqrt{\dfrac{x+3}{2}}\)
d, \(2\left(x^2-3x+2\right)=3\sqrt{x^3+8}\)
e, \(729x^4+8\sqrt{1-x^2}=36\)
f, \(7x^2-10x+14=5\sqrt{x^4+4}\)
g, \(x^3+3x^2-3\sqrt[3]{3x+5}=1-3x\)
h, \(\sqrt{4-3\sqrt{10-3x}}=x-2\)
i, \(\sqrt{x-1}+\sqrt{x^2-1}=\sqrt{x^2-5x+4}\)
a) \(\sqrt{1+x}-\sqrt{8-x}+\sqrt{\left(1+x\right)\left(8-x\right)}=3\)
đặt t \(=\sqrt{1+x}-\sqrt{8-x}\)
\(\Leftrightarrow t^2=1+x-2\sqrt{\left(1+x\right)\left(8-x\right)}+8-x\)
\(\Leftrightarrow\sqrt{\left(1+x\right)\left(8-x\right)}=\dfrac{9-t^2}{2}\)
pt \(\Rightarrow t+\dfrac{9-t^2}{2}=3\)
\(\Leftrightarrow t^2-2t-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=-1\\t=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{1+x}-\sqrt{8-x}=-1\\\sqrt{1+x}-\sqrt{8+x}=3\end{matrix}\right.\)
suy ra tìm đc x
câu b đặt t =\(3x^2+5x+8\)
ta có pt \(\Leftrightarrow\sqrt{t}-\sqrt{t-7}=1\)
\(\Rightarrow t=16\)
\(\Leftrightarrow3x^2+5x+8=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{8}{3}\end{matrix}\right.\)