\(\sqrt{4x}\) - \(\sqrt{9x}\) + \(\sqrt{16x}\) = 2
b, \(\sqrt{4x}\) + \(2\sqrt{16x}\) - \(\sqrt{25x}\) = 1,2
c, \(\sqrt{16\left(x-1\right)}\) - \(\sqrt{x-1}\) + \(\sqrt{49\left(x-1\right)}\) =5
giải pt:
a,\(\left(13-4x\right)\sqrt{2x-3}+\left(4x-3\right)\sqrt{5-2x}=2+8\sqrt{-4x^2+16x-15}\)
b,\(\left(9x-2\right)\sqrt{3x-1}+\left(10-9x\right)\sqrt{3-3x}-4\sqrt{-9x^2+12x-3}=4\)
c, \(\left(6x-5\right)\sqrt{x+1}-\left(6x+2\right)\sqrt{x-1}+4\sqrt{x^2-1}=4x-3\)
giải phương trình
1)\(\sqrt{9\left(x-1\right)}=21\)
2)\(\sqrt{1-x}+\sqrt{4-4x}-\dfrac{1}{3}\sqrt{16-16x}+5=0\)
3)\(\sqrt{2x}-\sqrt{50}=0\)
4)\(\sqrt{4x^2+4x+1}=6\)
5)\(\sqrt{\left(x-3\right)^2}=3-x\)
1) \(\sqrt[]{9\left(x-1\right)}=21\)
\(\Leftrightarrow9\left(x-1\right)=21^2\)
\(\Leftrightarrow9\left(x-1\right)=441\)
\(\Leftrightarrow x-1=49\Leftrightarrow x=50\)
2) \(\sqrt[]{1-x}+\sqrt[]{4-4x}-\dfrac{1}{3}\sqrt[]{16-16x}+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}+\sqrt[]{4\left(1-x\right)}-\dfrac{1}{3}\sqrt[]{16\left(1-x\right)}+5=0\)
\(\)\(\Leftrightarrow\sqrt[]{1-x}+2\sqrt[]{1-x}-\dfrac{4}{3}\sqrt[]{1-x}+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}\left(1+3-\dfrac{4}{3}\right)+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}.\dfrac{8}{3}=-5\)
\(\Leftrightarrow\sqrt[]{1-x}=-\dfrac{15}{8}\)
mà \(\sqrt[]{1-x}\ge0\)
\(\Leftrightarrow pt.vô.nghiệm\)
3) \(\sqrt[]{2x}-\sqrt[]{50}=0\)
\(\Leftrightarrow\sqrt[]{2x}=\sqrt[]{50}\)
\(\Leftrightarrow2x=50\Leftrightarrow x=25\)
1) \(\sqrt{9\left(x-1\right)}=21\) (ĐK: \(x\ge1\))
\(\Leftrightarrow3\sqrt{x-1}=21\)
\(\Leftrightarrow\sqrt{x-1}=7\)
\(\Leftrightarrow x-1=49\)
\(\Leftrightarrow x=49+1\)
\(\Leftrightarrow x=50\left(tm\right)\)
2) \(\sqrt{1-x}+\sqrt{4-4x}-\dfrac{1}{3}\sqrt{16-16x}+5=0\) (ĐK: \(x\le1\))
\(\Leftrightarrow\sqrt{1-x}+2\sqrt{1-x}-\dfrac{4}{3}\sqrt{1-x}+5=0\)
\(\Leftrightarrow\dfrac{5}{3}\sqrt{1-x}+5=0\)
\(\Leftrightarrow\dfrac{5}{3}\sqrt{1-x}=-5\) (vô lý)
Phương trình vô nghiệm
3) \(\sqrt{2x}-\sqrt{50}=0\) (ĐK: \(x\ge0\))
\(\Leftrightarrow\sqrt{2x}=\sqrt{50}\)
\(\Leftrightarrow2x=50\)
\(\Leftrightarrow x=\dfrac{50}{2}\)
\(\Leftrightarrow x=25\left(tm\right)\)
4) \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\left(ĐK:x\ge-\dfrac{1}{2}\right)\\2x+1=-6\left(ĐK:x< -\dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\left(tm\right)\\x=-\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)
5) \(\sqrt{\left(x-3\right)^2}=3-x\)
\(\Leftrightarrow\left|x-3\right|=3-x\)
\(\Leftrightarrow x-3=3-x\)
\(\Leftrightarrow x+x=3+3\)
\(\Leftrightarrow x=\dfrac{6}{2}\)
\(\Leftrightarrow x=3\)
1) => 9(x-1)=\(21^2\)
=> 9x-9=441
=> 9x=450
=> x=50
2)=>\(\sqrt{1-x}\) + \(\sqrt{4\left(1-x\right)}\)-\(\dfrac{1}{3}\sqrt{16\left(1-x\right)}\)+5=0
=>\(\sqrt{1-x}\)\(\left(1+2-\dfrac{1}{3}.4\right)\)+5=0
=>\(\dfrac{5}{3}\sqrt{1-x}\) +5=0
=>\(\sqrt{1-x}\)=-3
Phuong trinh vo nghiem
giải pt , \(\sqrt{x^4+4x^2}+\sqrt{x+x^2}=\sqrt{\left(x^2+\sqrt{x}\right)^2+9x^2}.\)
\(x=0\)
\(x^3=0\)
\(x^3=2.0.\sqrt{0}\)
\(x^3=2x\sqrt{x}\)
\(x^3=2x\sqrt{x}\)
\(4\left(x^3-2x\sqrt{x}\right)^2=0\)
\(4\left(x^6-4x^4\sqrt{x}+4x^2x\right)=0\)
\(4x^6-16x^4\sqrt{x}+16x^2x=0\)
\(4x^6+16x^3=16x^4\sqrt{x}\)
\(16x^4+4x^5+4x^6+16x^3=16x^4+4x^5+16x^4\sqrt{x}\)
\(4x^3\left(x+1\right)\left(x^2+4\right)=4\left(4x^4+4x^4\sqrt{x}+x^4.x\right)\)
\(4x^3\left(x+1\right)\left(x^2+4\right)=4\left(2x^2+x^2\sqrt{x}\right)^2\)
\(2\sqrt{2x^3\left(x+1\right)\left(x^2+4\right)}=2\left(2x^2+x^2\sqrt{x}\right)\)
\(x^4+x^2+4x^2+x+2\sqrt{2x^3\left(x+1\right)\left(x^2+4\right)}=2\left(2x^2+x^2\sqrt{x}\right)+x^4+x^2+4x^2+x\)
\(\left(\sqrt{x^4+4x^2}+\sqrt{x^2+x}\right)^2=\left(x^4+2x^2\sqrt{x}+x\right)+9x^2\)
\(\sqrt{x^4+4x^2}+\sqrt{x^2+x}=\sqrt{\left(x^2+\sqrt{x}\right)^2+9x^2}\)
vậy x=0 là nghiệm của pt =))
cho mk hỏi một chút là đây đích thực có phải lớp 1 ko ak?
giải phương trình
a)\(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\)
b)\(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\)
c)\(\sqrt{4x+20}+\sqrt{x+5}-\dfrac{1}{3}\sqrt{9x+45}=4\)
d)\(\dfrac{1}{3}\sqrt{2x}-\sqrt{8x}+\sqrt{18x}-10=2\)
a) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐK: \(x\ge1\))
\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow-2\sqrt{x-1}=-2\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{2}{2}\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\left(tm\right)\)
b) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\) (ĐK: \(x\ge-1\))
\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
\(\Leftrightarrow x+1=16\)
\(\Leftrightarrow x=15\left(tm\right)\)
Bài 1: Giải phương trình:
a, \(\dfrac{3}{4}\sqrt{4x}-\sqrt{4x}+5=\dfrac{1}{4}\sqrt{4x}\)
b,\(\sqrt{3-x}-\sqrt{27-9x}+1,25\sqrt{48-16x}=6\)
Bài 2: Cho biểu thức:
P=\(\left(\dfrac{2}{\sqrt{1+a}}+\sqrt{1-a}\right):\left(\dfrac{2}{1-a^2}+1\right)\) (với a\(\ge\)0; a\(\ne\)1)
a, Rút gọn P
b, Tính giá trị của P với a=\(\dfrac{24}{49}\)
c, Tìm a để P=2
Tôi cần gấp hai bài này vào chiều ngày 9 tháng 8 nên mong mọi người giúp đỡ ạ
a) ĐK: \(x\ge0\)
PT \(\Leftrightarrow\sqrt{4x}\left(\dfrac{3}{4}-1-\dfrac{1}{4}\right)+5=0\)
\(\Leftrightarrow2\sqrt{x}.\left(-\dfrac{1}{2}\right)+5=0\)
\(\Leftrightarrow x=25\) (thỏa)
Vậy \(x=25\)
b) Đk: \(x\le3\)
PT \(\Leftrightarrow\sqrt{3-x}-\sqrt{9\left(3-x\right)}+\dfrac{5}{4}\sqrt{16\left(3-x\right)}=6\)
\(\Leftrightarrow\sqrt{3-x}\left(1-\sqrt{9}+\dfrac{5}{4}.\sqrt{16}\right)=6\)
\(\Leftrightarrow\sqrt{3-x}=2\Leftrightarrow x=-1\) (thỏa)
Vậy \(x=-1\)
2:
a:
Sửa đề: \(P=\left(\dfrac{2}{\sqrt{1+a}}+\sqrt{1-a}\right):\left(\dfrac{2}{\sqrt{1-a^2}}+1\right)\)
\(P=\dfrac{2+\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{1+a}}:\dfrac{2+\sqrt{1-a^2}}{\sqrt{1-a^2}}\)
\(=\dfrac{2+\sqrt{1-a^2}}{\sqrt{1+a}}\cdot\dfrac{\sqrt{1-a^2}}{2+\sqrt{1-a^2}}=\sqrt{\dfrac{1-a^2}{1+a}}\)
\(=\sqrt{1-a}\)
b: Khi a=24/49 thì \(P=\sqrt{1-\dfrac{24}{49}}=\sqrt{\dfrac{25}{49}}=\dfrac{5}{7}\)
c: P=2
=>1-a=4
=>a=-3
1a (đkxđ:\(x\ge0\)) \(\Leftrightarrow\dfrac{-1}{2}.\sqrt{4x}+5=0\) \(\Leftrightarrow\sqrt{4x}=10\) \(\Leftrightarrow x=25\) (t/m)
b (đkxđ:\(x\le3\) ) \(\Leftrightarrow\sqrt{3-x}\left(1-3+1,25.4\right)=6\) \(\Leftrightarrow\sqrt{3-x}=2\) \(\Leftrightarrow x=-1\) (t/m)
Giải các phương trình sau:
a. \(\sqrt{25x+75}+3\sqrt{x-2}=2\sqrt{x-2}+\sqrt{9x-18}\)
b. \(\sqrt{\left(2x-1\right)^2}=4\)
c. \(\sqrt{\left(2x+1\right)^2}=3x-5\)
d. \(\sqrt{4x-12}-14\sqrt{\dfrac{x-2}{49}}=\sqrt{9x-18}+8\)
a) Ta có: \(\sqrt{25x+75}+3\sqrt{x-2}=2\sqrt{x-2}+\sqrt{9x-18}\)
\(\Leftrightarrow5\sqrt{x+3}+3\sqrt{x-2}=2\sqrt{x-2}+3\sqrt{x-2}\)
\(\Leftrightarrow\sqrt{25x+75}=\sqrt{4x-8}\)
\(\Leftrightarrow25x-4x=-8-75\)
\(\Leftrightarrow21x=-83\)
hay \(x=-\dfrac{83}{21}\)
b) Ta có: \(\sqrt{\left(2x-1\right)^2}=4\)
\(\Leftrightarrow\left|2x-1\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=4\\2x-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
c) Ta có: \(\sqrt{\left(2x+1\right)^2}=3x-5\)
\(\Leftrightarrow\left|2x+1\right|=3x-5\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=3x-5\left(x\ge-\dfrac{1}{2}\right)\\2x+1=5-3x\left(x< \dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3x=-5-1\\2x+3x=5-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\left(nhận\right)\\x=\dfrac{4}{5}\left(loại\right)\end{matrix}\right.\)
d) Ta có: \(\sqrt{4x-12}-14\sqrt{\dfrac{x-2}{49}}=\sqrt{9x-18}+8\)
\(\Leftrightarrow2\sqrt{x-3}-2\sqrt{x-2}=3\sqrt{x-2}+8\)
\(\Leftrightarrow2\sqrt{x-3}-5\sqrt{x-2}=8\)
\(\Leftrightarrow4\left(x-3\right)+25\left(x-2\right)-20\sqrt{x^2-5x+6}=8\)
\(\Leftrightarrow4x-12+25x-50-8=20\sqrt{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow20\sqrt{\left(x-2\right)\left(x-3\right)}=29x-70\)
\(\Leftrightarrow x^2-5x+6=\dfrac{\left(29x-70\right)^2}{400}\)
\(\Leftrightarrow x^2-5x+6=\dfrac{841}{400}x^2-\dfrac{203}{20}x+\dfrac{49}{4}\)
\(\Leftrightarrow\dfrac{-441}{400}x^2+\dfrac{103}{20}x-\dfrac{25}{4}=0\)
\(\Delta=\left(\dfrac{103}{20}\right)^2-4\cdot\dfrac{-441}{400}\cdot\dfrac{-25}{4}=-\dfrac{26}{25}\)(Vô lý)
vậy: Phương trình vô nghiệm
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
Giải phương trình:
a. \(3\sqrt{8x}-\sqrt{32x}+\sqrt{50x}=21\)
b. \(\sqrt{25x+50}+3\sqrt{4x+8}-2\sqrt{16x+32}=15\)
c. \(\sqrt{\left(x-2\right)^2}=12\)
d. \(\sqrt{x^2-6x+9}-3=5\)
e.\(\sqrt{\left(2x-1\right)^2}-x=3\)
f. \(\sqrt{3x-6}-x=-2\)
h. \(\sqrt{3-2x}-2=x\)
a.
ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow 6\sqrt{2x}-4\sqrt{2x}+5\sqrt{2x}=21$
$\Leftrightarrow 7\sqrt{2x}=21$
$\Leftrightarrow \sqrt{2x}=3$
$\Leftrightarrow 2x=9$
$\Leftrightarrow x=\frac{9}{2}$ (tm)
b.
ĐKXĐ: $x\geq -2$
PT $\Leftrightarrow \sqrt{25(x+2)}+3\sqrt{4(x+2)}-2\sqrt{16(x+2)}=15$
$\Leftrightarrow 5\sqrt{x+2}+6\sqrt{x+2}-8\sqrt{x+2}=15$
$\Leftrightarrow 3\sqrt{x+2}=15$
$\Leftrightarrow \sqrt{x+2}=5$
$\Leftrightarrow x+2=25$
$\Leftrightarrow x=23$ (tm)
c.
$\sqrt{(x-2)^2}=12$
$\Leftrightarrow |x-2|=12$
$\Leftrightarrow x-2=12$ hoặc $x-2=-12$
$\Leftrightarrow x=14$ hoặc $x=-10$
e.
PT $\Leftrightarrow |2x-1|-x=3$
Nếu $x\geq \frac{1}{2}$ thì $2x-1-x=3$
$\Leftrightarrow x=4$ (tm)
Nếu $x< \frac{1}{2}$ thì $1-2x-x=3$
$\Leftrightarrow x=\frac{-2}{3}$ (tm)
f.
ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow \sqrt{3(x-2)}-(x-2)=0$
$\Leftrightarrow \sqrt{x-2}(\sqrt{3}-\sqrt{x-2})=0$
$\Leftrightarrow \sqrt{x-2}=0$ hoặc $\sqrt{3}-\sqrt{x-2}=0$
$\Leftrightarrow x=2$ hoặc $x=5$ (tm)
h. ĐKXĐ: $x\leq \frac{3}{2}$
PT $\Leftrightarrow \sqrt{3-2x}=x+2$
\(\Rightarrow \left\{\begin{matrix} x+2\geq 0\\ 3-2x=(x+2)^2=x^2+4x+4\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq -2\\ x^2+6x+1=0\end{matrix}\right.\)
\(\Leftrightarrow x=-3+2\sqrt{2}\) (tm)
Vậy.......
\(\sqrt{25x+25}-\sqrt{16x+16}+\sqrt{9x+9}-\sqrt{4x+4}+\sqrt{x+1}=27\)
ĐKXĐ: \(x\ge-1\)
\(\sqrt{25\left(x+1\right)}-\sqrt{16\left(x+1\right)}+\sqrt{9\left(x+1\right)}-\sqrt{4\left(x+1\right)}+\sqrt{x+1}=27\)
\(\Leftrightarrow5\sqrt{x+1}-4\sqrt{x+1}+3\sqrt{x+1}-2\sqrt{x+1}+\sqrt{x+1}=27\)
\(\Leftrightarrow3\sqrt{x+1}=27\)
\(\Leftrightarrow\sqrt{x+1}=9\)
\(\Rightarrow x+1=81\)
\(\Rightarrow x=80\) (thỏa mãn)