tìm x
a)2(x2+1)=9x(\(\sqrt{x+2}\) +1)2
b)\(\sqrt{x}\) \(+\sqrt{9-x}-\sqrt{x\left(9-x\right)}\) =3
c) 2x-x2 +\(\sqrt{6x^2-12x}+7\) =0
help me please
Giải phương trình sau:
a) \(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)
b) \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
c) \(2x-x^2+\sqrt{6x^2-12x+7}=0\)
d) \(\left(x+1\right)\left(x+4\right)-3\sqrt{x^2+5x+2}=6\)
a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow x+5=4\)
hay x=-1
b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\Leftrightarrow\sqrt{x-1}=17\)
\(\Leftrightarrow x-1=289\)
hay x=290
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,\(\left(13-4x\right)\sqrt{2x-3}+\left(4x-3\right)\sqrt{5-2x}=2+8\sqrt{-4x^2+16x-15}\)
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 PT:
a) \(\dfrac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}.\)
b) \(\sqrt{4x-20}+3\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4.\)
c) \(2x-x^2+\sqrt{6x^2-12x+7}=0.\)
d) \(\left(x+1\right)\left(x+4\right)-3\sqrt{x^2+5x+2}=6.\)
a)
ĐKXĐ: \(x> \frac{-5}{7}\)
Ta có: \(\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)
\(\Rightarrow 9x-7=\sqrt{7x+5}.\sqrt{7x+5}=7x+5\)
\(\Rightarrow 2x=12\Rightarrow x=6\) (hoàn toàn thỏa mãn)
Vậy......
b) ĐKXĐ: \(x\geq 5\)
\(\sqrt{4x-20}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=4\)
\(\Leftrightarrow \sqrt{4}.\sqrt{x-5}+3\sqrt{\frac{1}{9}}.\sqrt{x-5}-\frac{1}{3}\sqrt{9}.\sqrt{x-5}=4\)
\(\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow 2\sqrt{x-5}=4\Rightarrow \sqrt{x-5}=2\Rightarrow x-5=2^2=4\Rightarrow x=9\)
(hoàn toàn thỏa mãn)
Vậy..........
c) ĐK: \(x\in \mathbb{R}\)
Đặt \(\sqrt{6x^2-12x+7}=a(a\geq 0)\Rightarrow 6x^2-12x+7=a^2\)
\(\Rightarrow 6(x^2-2x)=a^2-7\Rightarrow x^2-2x=\frac{a^2-7}{6}\)
Khi đó:
\(2x-x^2+\sqrt{6x^2-12x+7}=0\)
\(\Leftrightarrow \frac{7-a^2}{6}+a=0\)
\(\Leftrightarrow 7-a^2+6a=0\)
\(\Leftrightarrow -a(a+1)+7(a+1)=0\Leftrightarrow (a+1)(7-a)=0\)
\(\Rightarrow \left[\begin{matrix} a=-1\\ a=7\end{matrix}\right.\) \(\Rightarrow a=7\) vì \(a\geq 0\)
\(\Rightarrow 6x^2-12x+7=a^2=49\)
\(\Rightarrow 6x^2-12x-42=0\Leftrightarrow x^2-2x-7=0\)
\(\Leftrightarrow (x-1)^2=8\Rightarrow x=1\pm 2\sqrt{2}\)
(đều thỏa mãn)
Vậy..........
d)
ĐKXĐ: \(x^2+5x+2\ge 0\)
\((x+1)(x+4)-3\sqrt{x^2+5x+2}=6\)
\(\Leftrightarrow (x^2+5x+4)-3\sqrt{x^2+5x+2}=6\)
Đặt \(\sqrt{x^2+5x+2}=a(a\geq 0)\Rightarrow x^2+5x+2=a^2\)
PT trở thành:
\(a^2+2-3a=6\)
\(\Leftrightarrow a^2-3a-4=0\Leftrightarrow (a-4)(a+1)=0\)
\(\Rightarrow a=4\) vì \(a\geq 0\)
\(\Rightarrow x^2+5x+2=a^2=16\)
\(\Rightarrow x^2+5x-14=0\Leftrightarrow (x-2)(x+7)=0\)
\(\Rightarrow \left[\begin{matrix} x=2\\ x=-7\end{matrix}\right.\) (đều thỏa mãn)
Vậy................
- Giải giúp mình những câu này với~
a) \(^{x2+2\sqrt{3x}+\sqrt{3}^2}=\left(x+\sqrt{3}\right)^2\))^2
b)(9x^2 -12x +4) - (y+2)^2
c)x^2-2x+2
b)9x^2 -6x +5
c)30-20x+4x2
Lớp 9 lẫn 7 luôn nhé
a và b chắc của lớp 9 nhỉ
\(x^2-2x+2=x^2-x-x+2\)
\(=x\left(x-1\right)-\left(x-1\right)+1\)
\(=\left(x-1\right)^2+1\)
\(9x^2-6x+5=9\left(x^2-\frac{2}{3}x+\frac{5}{9}\right)\)
\(=9\left(x^2-\frac{1}{3}x-\frac{1}{3}x+\frac{5}{9}\right)\)
\(=9\left(x^2-\frac{1}{3}x-\frac{1}{3}x+\frac{1}{9}+\frac{4}{9}\right)\)
\(=9\left[x\left(x-\frac{1}{3}\right)-\frac{1}{3}\left(x-\frac{1}{3}\right)+\frac{4}{9}\right]\)
\(=9\left[\left(x-\frac{1}{3}\right)^2+\frac{4}{9}\right]\)
\(=9\left(x-\frac{1}{3}\right)^2+4\)
Cái kia tương tự.
2. Giải PT:
a) \(\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}.\)
b) \(\sqrt{4x-20}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=4.\)
c) \(2x-x^2+\sqrt{6x^2-12x+7}=0.\)
d) \(\left(x+1\right)\left(x+4\right)-3\sqrt{x^2+5x+2}=6.\)
\(a,\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)\(ĐKXĐ:x\ge-\frac{5}{7}\)
\(\Leftrightarrow9x-7=7x+5\)
\(\Leftrightarrow9x-7x=5+7\)
\(\Leftrightarrow2x=12\)
\(\Leftrightarrow x=6\)
\(b,\sqrt{4x-20}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=4\)
\(\Leftrightarrow\sqrt{4\left(x-5\right)}+3.\frac{\sqrt{x-5}}{\sqrt{9}}-\frac{1}{3}\sqrt{9\left(x-5\right)}=4\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}\left(2+1-1\right)=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\)
\(\Leftrightarrow x=9\)
giải phương trình
a, \(\sqrt{4x-20}+3\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
b, \(2x-x^2+\sqrt{6x^2-12x+7}=0\)
c, \(\dfrac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)
Tất cả đều có điều kiện \(x\ge0\)
a,\(\sqrt{x^2-6x+9}+x=11\)
b,\(\sqrt{3x^2-4x+3=1-2x}\)
c,\(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\)
d,\(\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\)
Đề yc giải pt à em?
Câu b bạn có bị lỗi dấu căn không mà sao nó kéo dài cả 2 vế pt vậy :v
\(a,\sqrt{x^2-6x+9}+x=11\\ \Leftrightarrow\sqrt{\left(x-3\right)^2}=11-x\)
\(\Leftrightarrow\left|x-3\right|=11-x\\ TH_1:x\ge3\\ x-3=11-x\\ \Leftrightarrow2x=14\\ \Leftrightarrow x=7\left(tm\right)\)
\(TH_2:x< 3\\ -x+3=11-x\\ \Leftrightarrow-x+x=11-3\\ \Leftrightarrow0=8\left(VL\right)\)
Vậy \(S=\left\{7\right\}\)
\(c,\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\) \(\left(dk:x\ge-1\right)\)
\(\Leftrightarrow\sqrt{4^2}.\sqrt{\left(x+1\right)}-\sqrt{3^2}.\sqrt{\left(x+1\right)}=4\left(1\right)\)
Đặt \(a=\sqrt{x+1}\left(a\ge0\right)\)
Pt trở thành : \(4a-3a=4\Leftrightarrow a=4\left(tmdk\right)\)
\(\Rightarrow\sqrt{x+1}=4\\ \Rightarrow\left(\sqrt{x+1}\right)^2=16\\ \Rightarrow\left|x+1\right|=16\)
\(TH_1:x\ge-1\\ x+1=16\Leftrightarrow x=15\left(tm\right)\\ TH_2:x< -1\\ -x-1=16\Leftrightarrow x=-17\left(tm\right)\)
Nhưng loại TH2 vì dk ban đầu là \(x\ge-1\)
Vậy \(S=\left\{15\right\}\)
\(d,\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\left(dk:x\ge-1\right)\\ \Leftrightarrow\sqrt{9}.\sqrt{x+1}+\sqrt{4}.\sqrt{x+1}-\sqrt{x+1}=0\)
Đặt \(\sqrt{x+1}=a\left(a\ge0\right)\)
Tới đây bạn làm tương tự câu c nha.
Bài 1: giải p.trình
a,\(\sqrt{x^2-4x+4}=1\)
b,\(\sqrt{1-4x+4x^2}=5\)
c,\(\sqrt{a\left(1-2x+x^2\right)}-6=0\)
d,\(\sqrt{9x^2}=2x+1\)
e,\(\sqrt{9-6x+x^2}=x\)
a, ĐKXĐ: \(x^2-4x+4\ge0\Rightarrow\left(x-2\right)^2\ge0\left(luônđúng\right)\)
\(\sqrt{x^2-4x+4}=1\\ \Rightarrow x-2=1\\ \Rightarrow x=3\)
b,\(ĐKXĐ:1-4x+4x^2\ge0\Rightarrow\left(1-2x\right)^2\ge0\left(luônđúng\right)\)
\(\sqrt{1-4x+4x^2}=5\\ \Rightarrow\left|1-2x\right|=5\\ \Rightarrow\left[{}\begin{matrix}1-2x=5\\1-2x=-5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
d, ĐKXĐ: \(\left\{{}\begin{matrix}9x^2\ge0\\2x+1\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge0\\x\ge-\dfrac{1}{2}\end{matrix}\right.\Rightarrow x\ge0\)
\(\sqrt{9x^2}=2x+1\\ \Rightarrow\left|3x\right|=2x+1\\ \Rightarrow\left[{}\begin{matrix}3x=2x+1\\3x=-2x+1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
c, ĐKXĐ: \(1-2x+x^2\ge0\Rightarrow\left(1-x\right)^2\ge0\left(luônđúng\right)\)
\(\sqrt{1-2x+x^2}-6=0\\ \Rightarrow\left|1-x\right|=6\\ \Rightarrow\left[{}\begin{matrix}1-x=-6\\1-x=6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=7\\x=-5\end{matrix}\right.\)
e, \(\left\{{}\begin{matrix}9-6x+x^2\ge0\\x\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left(3-x\right)^2\ge0\left(luônđúng\right)\\x\ge0\end{matrix}\right.\)\(\Rightarrow x\ge0\)
\(\sqrt{9-6x+x^2}=x\\ \Rightarrow\left|3-x\right|=x\\ \Rightarrow\left[{}\begin{matrix}3-x=-x\\3-x=x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}3=0\left(vôlí\right)\\x=1,5\end{matrix}\right.\)
a) \(\sqrt{x^2-4x+4}=1\)
\(\Leftrightarrow\sqrt{\left(x-2\right)^2}=1\Leftrightarrow\left|x-2\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
b) \(\sqrt{1-4x+4x^2}=5\)
\(\Leftrightarrow\sqrt{\left(1-2x\right)^2}=5\Leftrightarrow\left|1-2x\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}1-2x=5\\1-2x=-5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
c) \(\sqrt{x\left(1-2x+x^2\right)}-6=0\)
\(\Leftrightarrow\left(\sqrt{x\left(1-x\right)^2}\right)^2=36\Leftrightarrow x\left(1-x\right)^2=36\)
\(\Leftrightarrow x-2x^2+x^3-36=0\)
\(\Leftrightarrow\left(x-4\right)\left(x^2+2x+9\right)=0\)
\(\Leftrightarrow x=4\)(do \(x^2+2x+9=\left(x+1\right)^2+8>0\))
d) \(\sqrt{9x^2}=2x+1\)
\(\Leftrightarrow3\left|x\right|=2x+1\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=2x+1\\-3x=2x+1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{5}\end{matrix}\right.\)
e) \(\sqrt{9-6x+x^2}=x\left(1\right)\left(đk:x\ge0\right)\)
\(\Leftrightarrow\sqrt{\left(3-x\right)^2}=x\Leftrightarrow\left|3-x\right|=x\)
TH1: \(0\le x\le3\)
\(\left(1\right)\Leftrightarrow3-x=x\Leftrightarrow x=\dfrac{3}{2}\)
TH2: \(x>3\)
\(\left(1\right)\Leftrightarrow x-3=x\Leftrightarrow-3=0\left(vn\right)\)