√x-2 + √4x-8 + √9x-18 = √25x-50+9
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)
Bài 1: Tìm x, biết
a)\(2\sqrt{9x-27}-\dfrac{1}{5}\sqrt{25x-75}-\dfrac{1}{7}\sqrt{49x-147}=20\)
b) \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)
c)\(\sqrt{16x-16}-\sqrt{9x-9}+\sqrt{4x-4}+\sqrt{x-1}=8\)
d) \(\sqrt{x+2\sqrt{x-1}}-\sqrt{x-2\sqrt{x-1}}=2\)
a) Ta có: \(2\sqrt{9x-27}-\dfrac{1}{5}\sqrt{25x-75}-\dfrac{1}{7}\sqrt{49x-147}=20\)
\(\Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)
\(\Leftrightarrow4\sqrt{x-3}=20\)
\(\Leftrightarrow x-3=25\)
hay x=28
b) Ta có: \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)
\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)
\(\Leftrightarrow2\sqrt{x+2}=6\)
\(\Leftrightarrow x+2=9\)
hay x=7
Tìm x:
\(\sqrt{9x+18}-0,5\sqrt{4x+8}-\dfrac{4}{5}\sqrt{25x+50}=-10\)
ĐK: \(x+2\ge0\Leftrightarrow x\ge-2\)
\(3\sqrt{x+2}-\sqrt{x+2}-4\sqrt{x+2}=-10\)
\(-2\sqrt{x+2}=-10\)
\(\sqrt{x+2}=5\)
\(\left\{{}\begin{matrix}5\ge0\left(ld\right)\\x+2=25\end{matrix}\right.\)\(\Leftrightarrow x=23\left(n\right)\)
Giải các phương trình sau:
a. \(\sqrt{25x+75}+2\sqrt{9x+27}=5\sqrt{x+3}+18\)
b. \(\sqrt{4x-8}-14\sqrt{\dfrac{x-2}{49}}=\sqrt{9x-18}+8\)
a) Ta có: \(\sqrt{25x+75}+2\sqrt{9x+27}=5\sqrt{x+3}+18\)
\(\Leftrightarrow5\sqrt{x+3}+6\sqrt{x+3}-5\sqrt{x+3}=18\)
\(\Leftrightarrow\sqrt{x+3}=3\)
\(\Leftrightarrow x+3=9\)
hay x=6
b) Ta có: \(\sqrt{4x-8}-14\sqrt{\dfrac{x-2}{49}}=\sqrt{9x-18}+8\)
\(\Leftrightarrow2\sqrt{x-2}-2\sqrt{x-2}-3\sqrt{x-2}=8\)
\(\Leftrightarrow-3\sqrt{x-2}=8\)(Vô lý)
Tìm x biết
2,\(\frac{1}{5}\sqrt{25x+50}-5\sqrt{x+2}+\sqrt{9x+18}+9=0\)
3,\(\sqrt{x^2-4x+4}=7x-1\)
2) \(\frac{1}{5}\sqrt{25x+50}-5\sqrt{x+2}+\sqrt{9x+18}+9=0\)
\(\frac{1}{5}\sqrt{25\left(x+2\right)}-5\sqrt{x+2}+\sqrt{9x+18}+9=0\)
\(\frac{1}{5}.\sqrt{25}.\sqrt{x+2}-5\sqrt{x+2}+\sqrt{9x+18}+9=0\)
\(\frac{1}{5}.5\sqrt{x+2}-5\sqrt{x+2}+\sqrt{9x+18}+9=0\)
\(\frac{1}{5}.5\sqrt{x+2}-5\sqrt{x+2}+\sqrt{9\left(x+2\right)}+9=0\)
\(\frac{1}{5}.5\sqrt{x+2}-5\sqrt{x+2}+\sqrt{9}.\sqrt{x+2}+9=0\)
\(\frac{1}{5}.5\sqrt{x+2}-5\sqrt{x+2}+3\sqrt{x+2}+9=0\)
\(\sqrt{x+2}-5\sqrt{x+2}+3\sqrt{x+2}+9=0\)
\(-\sqrt{x+2}=-9\)
\(x+2=81\)
\(\Rightarrow x=79\)
3) \(\sqrt{x^2-4x+4}=7x-1\)
\(\sqrt{x^2-2.x.2+2^2}=7x-1\)
\(\sqrt{\left(x-2\right)^2}=7x-1\)
\(x-2=7x-1\)
\(-2=7x-1-x\)
\(-2+1=7x-x\)
\(-1=6x\)
\(-\frac{1}{6}=x\)
\(\Rightarrow x=-\frac{1}{6}\)
Phân tích đa thức thành nhân tử
a) x^8+ x^7+ 1
b) 8x^4- 4x^3+ 2x^2+ 9x- 45
c) 16x^4- 16x^3+ 7x^2+ 9x- 9
d) 24x^4- 4x^3- 158x^2+ 25x+ 50
e) 2x^3- x^2- 15x- 22x+ 8
giải phương trình sau:
\(\sqrt{25x-50}-\dfrac{\sqrt{x-2}-1}{2}=8\sqrt{\dfrac{9x-18}{16}}\)
\(Đk:x\ge2\\ PT\Leftrightarrow\dfrac{10\sqrt{x-2}-\sqrt{x-2}+1}{2}=6\sqrt{x-2}\\ \Leftrightarrow9\sqrt{x-2}+1=12\sqrt{x-2}\\ \Leftrightarrow\sqrt{x-2}=\dfrac{1}{3}\Leftrightarrow x-2=\dfrac{1}{9}\\ \Leftrightarrow x=\dfrac{19}{9}\left(tm\right)\)
Bài 1: GIẢI PHƯƠNG TRÌNH
a) \(\sqrt{49-28x-4x^2}-5=0\)
b) \(\frac{1}{2}\sqrt{x-2}-4\sqrt{\frac{4x-8}{9}}-5=0\)
c) \(\sqrt{x^2-4x+4}=7x-1\)
d) \(\frac{1}{5}\sqrt{25x+50}-5\sqrt{x+2}+\sqrt{9x+18}+9=0\)
a.
\(DK:49-28x-4x^2\ge0\)
PT\(\Leftrightarrow\sqrt{49-28x-4x^2}=5\)
\(\Leftrightarrow49-28x-4x^2=25\)
\(\Leftrightarrow4x^2+28x-24=0\)
\(\Leftrightarrow x^2+7x-6=0\)
Ta co:
\(\Delta=7^2-4.1.\left(-6\right)=73>0\)
\(\Rightarrow\hept{\begin{cases}x_1=\frac{-7+\sqrt{73}}{2}\left(n\right)\\x_2=\frac{-7-\sqrt{73}}{2}\left(n\right)\end{cases}}\)
Vay nghiem cua PT la \(\hept{\begin{cases}x_1=\frac{-7+\sqrt{73}}{2}\\x_2=\frac{-7-\sqrt{73}}{2}\end{cases}}\)
\(\dfrac{1}{5}\)\(\sqrt{25x+50}\) - 5\(\sqrt{x+2}\) + \(\sqrt{9x+18}\) + 9 = 0 ( Giải phương trình sau )
Mong mng giúp đỡ ạ!
\(\dfrac{1}{5}\sqrt[]{25x+50}-5\sqrt[]{x+2}+\sqrt[]{9x+18}+9=0\)
\(\Leftrightarrow\dfrac{1}{5}\sqrt[]{25\left(x+2\right)}-5\sqrt[]{x+2}+\sqrt[]{9\left(x+2\right)}+9=0\)
\(\Leftrightarrow\dfrac{1}{5}.5\sqrt[]{x+2}-5\sqrt[]{x+2}+3\sqrt[]{x+2}+9=0\)
\(\Leftrightarrow\sqrt[]{x+2}-5\sqrt[]{x+2}+3\sqrt[]{x+2}+9=0\)
\(\Leftrightarrow\sqrt[]{x+2}\left(1-5+3\right)+9=0\)
\(\Leftrightarrow-\sqrt[]{x+2}+9=0\)
\(\Leftrightarrow\sqrt[]{x+2}=9\)
\(\Leftrightarrow x+2=81\)
\(\Leftrightarrow x=79\)