\(\sqrt{x^2+4\sqrt{x-4}}=5\)
giúp mị vs: fb Auoland z'yanh
Sử dụng hằng đẳng thức \(\sqrt{A^2}\)= \(\)IAI để giải pt:
a) \(\sqrt{9-12x+4x^2}\)= 4 + x
b) \(\sqrt{4-4x+x^2}\)= ( x - 1 )2 + x - 6
Mọi người ơi giúp mị vs gấp lắm !!
a, \(\Leftrightarrow\sqrt{\left(3-2x\right)^2=4+x}\)
\(\Leftrightarrow\left|3-2x\right|=4+x\)
\(\Leftrightarrow\orbr{\begin{cases}3-2x=4+x\\3-2x=-4-x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x=-1\\x=7\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=7\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\orbr{\begin{cases}x=\sqrt{7}\\x=-\sqrt{7}\end{cases}}\\\left(x-3\right)\left(x-1\right)=0\end{cases}}\)
2) giải pt
3) \(\sqrt{4x+1}=x+1\)
4) \(2\sqrt{x-1}+\dfrac{1}{3}\sqrt{9x-9}=15\)
5) \(\sqrt{4x^2-12x+9}=7\)
6) \(5\sqrt{9x-9}-\sqrt{4x-4}-\sqrt{x-1}=36\)
giúp mk vs ah
3: Ta có: \(\sqrt{4x+1}=x+1\)
\(\Leftrightarrow x^2+2x+1=4x+1\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=2\left(nhận\right)\end{matrix}\right.\)
4: Ta có: \(2\sqrt{x-1}+\dfrac{1}{3}\sqrt{9x-9}=15\)
\(\Leftrightarrow3\sqrt{x-1}=15\)
\(\Leftrightarrow x-1=25\)
hay x=26
5: Ta có: \(\sqrt{4x^2-12x+9}=7\)
\(\Leftrightarrow\left|2x-3\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
ai có thể giúp mình giải bài này vs đc không mình đang cần rất gấp (làm chi tiết hộ mình nhé, xin cảm ơn)
Bài 4:
a, \(\sqrt{3x+4}-\sqrt{2x+1}=\sqrt{x+3}\)
b, \(\sqrt{2x-5}+\sqrt{x+2}=\sqrt{2x+1}\)
c, \(\sqrt{x+4}-\sqrt{1-x}=\sqrt{1-2x}\)
d,\(\sqrt{x+9}=5-\sqrt{2x+4}\)
Bài 5:
a, \(\sqrt{x+4\sqrt{x}+4}=5x+2\)
b, \(\sqrt{x^2-2x+1}+\sqrt{x^2+4x+4}=4\)
c, \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=2\)
d,\(\sqrt{x-2+\sqrt{2x-5}}+\sqrt{x+2+3\sqrt{2x-5}}=7\sqrt{2}\)
Ví Dụ 1:
a, \(\sqrt{2x-1}=\sqrt{2}-1\)
b, \(\sqrt{x+5}=3-\sqrt{2}\)
c, \(\sqrt{3x^2}-\sqrt{12}=0\)
d, \(\sqrt{2}\left(x-1\right)-\sqrt{50}=0\)
Vd1:
d) Ta có: \(\sqrt{2}\left(x-1\right)-\sqrt{50}=0\)
\(\Leftrightarrow\sqrt{2}\left(x-1-5\right)=0\)
\(\Leftrightarrow x=6\)
2) giải pt
a) \(\sqrt{4-2x}=5\)
b) \(\sqrt{25\left(x+1\right)}+\sqrt{9x+9}=16\)
c) \(\sqrt{4x^2+12x+9}=4\)
giúp mk vs ạ mk cần gấp
a) ĐKXĐ: x <= 2
pt --> 4 - 2x = 25 <=> x = -21/2 (thỏa)
b) ĐKXĐ: x >= -1
pt <=> 8sqrt(x + 1)=16 <=> sqrt(x+1)=2 --> x + 1 = 4 <=> x = 3
(2) giải pt:
a) \(\sqrt{4-2x}=5\)
b) \(\sqrt{25\left(x+1\right)}+\sqrt{9x+9}=16\)
\(\sqrt{4x^2+12x+9}=4\)
giúp mk vs ạ mai mk hc rồi
a, ĐKXĐ: \(x\le2\)
\(\sqrt{4-2x}=5\\ \Leftrightarrow4-2x=25\\ \Leftrightarrow2x=-21\\ \Leftrightarrow x=-10,5\left(tm\right)\)
b, ĐKXĐ: \(x\ge-1\)
\(\sqrt{25\left(x+1\right)}+\sqrt{9x+9}=16\\ \Leftrightarrow5\sqrt{x+1}+\sqrt{9\left(x+1\right)}=16\\ \Leftrightarrow5\sqrt{x+1}+3\sqrt{x+1}=16\\ \Leftrightarrow8\sqrt{x+1}=16\\ \Leftrightarrow\sqrt{x+1}=2\\ \Leftrightarrow x+1=4\\ \Leftrightarrow x=3\)
c, \(\sqrt{4x^2+12x+9}=4\Leftrightarrow4x^2+12x+9=16\\ \Leftrightarrow4x^2+12x-7=0\\ \Leftrightarrow\left(4x^2-2x\right)+\left(14x-7\right)=0\\ \Leftrightarrow2x\left(2x-1\right)+7\left(2x-1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
a: \(\Leftrightarrow4-2x=25\)
hay \(x=-\dfrac{21}{2}\)
c: \(\Leftrightarrow\left|2x+3\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=4\\2x+3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
Mk đag cần gấp mn giúp mk vs ạ !
Câu 1 Tìm x , biết
a)\(\sqrt{4\text{x}^2+4\text{x}+1}=6\)
b)\(\sqrt{4\text{x}^2-4\sqrt{7}x+7=\sqrt{7}}\)
c\(\sqrt{x^2+2\sqrt{3}x+3}=2\sqrt[]{3}\)
d)\(\sqrt{\left(x-3\right)^2}=9\)
a) \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left(2x+1\right)^2=6^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
b) \(\sqrt{4x^2-4\sqrt{7}x+7}=\sqrt{7}\)
\(\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)
\(\Leftrightarrow\left(2x-\sqrt{7}\right)^2=\left(\sqrt{7}\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt[]{7}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)
a) \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
b) \(pt\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)
\(\Leftrightarrow\left|2x-\sqrt{7}\right|=\sqrt{7}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)
c) \(PT\Leftrightarrow\sqrt{\left(x+\sqrt{3}\right)^2}=2\sqrt{3}\)
\(\Leftrightarrow\left|x+\sqrt{3}\right|=2\sqrt{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\sqrt{3}=2\sqrt{3}\\x+\sqrt{3}=-2\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}\\x=-3\sqrt{3}\end{matrix}\right.\)
d) \(pt\Leftrightarrow\left|x-3\right|=9\Leftrightarrow\left[{}\begin{matrix}x-3=-9\\x-3=9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=12\end{matrix}\right.\)
giải pt:
a) \(4\sqrt{x-2}+\sqrt{9x-18}-\sqrt{\dfrac{x-2}{4}}=26\)
b) \(3x+\sqrt{4x^2-8x+4}=1\)
c) \(\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=7\)
giúp mk vs ạ mk cần gấp
giải pt:
a) \(4\sqrt{x-2}+\sqrt{9x-18}-\sqrt{\dfrac{x-2}{4}}=26\)
b) \(3x+\sqrt{4x^2-8x+4}=1\)
c) \(\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=7\)
giúp mk vs ạ mk cần gấp
a,ĐKXĐ:\(x\ge2\)
\(4\sqrt{x-2}+\sqrt{9x-18}-\sqrt{\dfrac{x-2}{4}}=26\\ \Leftrightarrow4\sqrt{x-2}+3\sqrt{x-2}-\dfrac{\sqrt{x-2}}{2}=26\\ \Leftrightarrow8\sqrt{x-2}+6\sqrt{x-2}-\sqrt{x-2}=52\\ \Leftrightarrow13\sqrt{x-2}=52\\ \Leftrightarrow\sqrt{x-2}=4\\ \Leftrightarrow x-2=16\\ \Leftrightarrow x=18\left(tm\right)\)
b,ĐKXĐ:\(x\in R\)
\(3x+\sqrt{4x^2-8x+4}=1\\ \Leftrightarrow2\sqrt{x^2-2x+1}=1-3x\\ \Leftrightarrow\left|x-1\right|=\dfrac{1-3x}{2}\\ \Leftrightarrow\left[{}\begin{matrix}x-1=\dfrac{1-3x}{2}\\x-1=\dfrac{3x-1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x-2=1-3x\\2x-2=3x-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
c, ĐKXĐ:\(x\ge0\)
\(\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=7\\ \Leftrightarrow\sqrt{x}\left(2\sqrt{x}+1\right)-2\left(2\sqrt{x}+1\right)=7\\ \Leftrightarrow2x+\sqrt{x}-4\sqrt{x}-2=7\\ \Leftrightarrow2x-3\sqrt{x}-9=0\\ \Leftrightarrow\left(2x+3\sqrt{x}\right)-\left(6\sqrt{x}+9\right)=0\\ \Leftrightarrow\sqrt{x}\left(2\sqrt{x}+3\right)-3\left(2\sqrt{x}+3\right)=0\\ \Leftrightarrow\left(\sqrt{x}-3\right)\left(2\sqrt{x}+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\\2\sqrt{x}=-3\left(vô.lí\right)\end{matrix}\right.\\ \Leftrightarrow x=9\left(tm\right)\)
1) giải phương trình
a)\(\sqrt{x+9}=7\)
b) \(\sqrt{x-4}=4-x\)
c) \(\sqrt{x^2-12x+36}=81\)
giúp mk vs ah
â) \(\sqrt{x+9}=7\\ \Rightarrow x+9=49\\ \Rightarrow x=40\)
b) \(\sqrt{x-4}=4-x\\ \Rightarrow x-4=16-8x+x^2\\ \Rightarrow x^2-9x+20=0\\ \Rightarrow\left(x-4\right)\left(x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
c) \(\sqrt{x^2-12x+36}=81\\ \Rightarrow x-6=81\\ \Rightarrow x=87\)
a: Ta có: \(\sqrt{x+9}=7\)
\(\Leftrightarrow x+9=49\)
hay x=40
b: Ta có: \(\sqrt{x-4}=4-x\)
\(\Leftrightarrow\left(x-4\right)^2-\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=5\left(loại\right)\end{matrix}\right.\)
c: Ta có: \(\sqrt{x^2-12x+36}=81\)
\(\Leftrightarrow\left|x-6\right|=81\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=81\\x-6=-81\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=87\\x=-75\end{matrix}\right.\)