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a)\(\sqrt{x+9}=7\)
Đk:\(x\ge-9\).Bình phương 2 vế của pt ta có:
\(\sqrt{\left(x+9\right)^2}=7^2\)\(\Leftrightarrow x+9=49\Leftrightarrow x=40\)
b)\(\sqrt{x^2-12x+36}=81\)
Đk:\(x\ge6\)
\(\Leftrightarrow\sqrt{\left(x-6\right)^2}=81\)
\(\Leftrightarrow x-6=81\Leftrightarrow x=87\)
c)\(\sqrt{x-1}=4\)
Đk:\(x\ge1\).Bình phương 2 vế của pt ta có:
\(\sqrt{\left(x-1\right)^2}=4^2\)
\(\Leftrightarrow x-1=16\Leftrightarrow x=17\)
c: \(\Leftrightarrow\sqrt{4x^2\left(x+2\right)}=3x+1\)
\(\Rightarrow\left\{{}\begin{matrix}4x^2\left(x+2\right)=9x^2+6x+1\\x>=-\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x^3+8x^2-9x^2-6x-1=0\\x>=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x^3-x^2-6x-1=0\\x>=-\dfrac{1}{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x^3+4x^2-5x^2-5x-x-1=0\\x>=-\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)\left(4x^2-5x-1\right)=0\\x>=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow x=\dfrac{5\pm\sqrt{41}}{8}\)
a: \(\Leftrightarrow\sqrt{5+\sqrt{x-1}}=6-x\)
\(\Leftrightarrow5+\sqrt{x-1}=x^2-12x+36\) và x<=6
=>\(\sqrt{x-1}=x^2-12x+31\) và x<=6
=>x-1=(x^2-12x+22+11)^2
=>\(x\in\varnothing\)
ĐKXĐ x>0
Chia cả 2 vế của pt cho \(\sqrt{x}\ne0\),ta được
\(12+\sqrt{\frac{x-1}{x}}=\frac{2}{x}+\sqrt{\frac{169x-65}{x}}\)
\(\Rightarrow12-\frac{2}{x}+\sqrt{1-\frac{1}{x}}=\sqrt{65\left(1-\frac{1}{x}\right)+104}\)(2)
Đặt \(\sqrt{1-\frac{1}{x}}=a\)(\(a\ge0\)),khi đó pt (1) trở thành
\(2a^2+10+a=\sqrt{65a^2+104}\)
\(\Leftrightarrow\left(2a^2+a+10\right)^2=65a^2+104\)
\(\Leftrightarrow\left(a-1\right)^2\left(a^2+3a-1\right)=0\)
Đến đây bn tự giải tiếp nhé
Trần Hữu Ngọc Minh xem tôi làm có đúng ko?
Giải:
a, \(\sqrt{2}.x-\sqrt{50}=0\)
\(\Leftrightarrow\sqrt{2}.x=\sqrt{50}\Leftrightarrow\sqrt{2}.x=\sqrt{25.2}\)
\(\Leftrightarrow\sqrt{2}.x=\sqrt{25}.\sqrt{2}\Leftrightarrow\sqrt{2}.x=5\sqrt{2}\)
\(\Leftrightarrow x=5\)
c, \(\sqrt{3}.x^2-\sqrt{12}=0\)
\(\Leftrightarrow\sqrt{3}.x^2=\sqrt{12}\)
\(\Leftrightarrow\sqrt{3}.x^2=\sqrt{4.3}\)
\(\Leftrightarrow\sqrt{3}.x^2=\sqrt{4}.\sqrt{3}\)
\(\Leftrightarrow\sqrt{3}.x^2=2\sqrt{3}\)
\(\Leftrightarrow x^2=2\)
\(\Leftrightarrow x=\pm\sqrt{2}\)
d, \(\frac{x^2}{\sqrt{5}}-\sqrt{20}=0\)
\(\Leftrightarrow\frac{x^2}{\sqrt{5}}=\sqrt{20}\)
\(\Leftrightarrow x^2=\sqrt{5}.\sqrt{20}\)
\(\Leftrightarrow x^2=\sqrt{100}\)
\(\Leftrightarrow x=\pm10\)
a, \(\sqrt{\left(x-1\right)^2}=5\Rightarrow\left(x-1\right)=\left\{5;-5\right\}\Leftrightarrow\hept{\begin{cases}x-1=5\Rightarrow x=6\\x-1=-5\Rightarrow x=-4\end{cases}}\)
b,\(3+\sqrt{x}=5\Rightarrow\sqrt{x}=2\Rightarrow x=4\)
c,\(\sqrt{x^2-2x+1}=x-1\Rightarrow\sqrt{\left(x-1\right)^2}=x-1\Rightarrow x-1=\left\{x-1;-\left(x-1\right)\right\}\)
\(\Leftrightarrow\hept{\begin{cases}x-1=x-1\Rightarrow x\in R\\x-1=-\left(x-1\right)\Rightarrow x-1=-x+1\Rightarrow x+x=1+1\Rightarrow2x=2\Rightarrow x=1\end{cases}}\)
Vậy x = 1
d, \(\sqrt{x^2-10x+25}=x+3\Rightarrow\sqrt{\left(x-5\right)^2}=x+3\Rightarrow x-5=\left\{x+3;-\left(x+3\right)\right\}\)
\(\Leftrightarrow\hept{\begin{cases}x-5=x+3\Rightarrow x-x=3+5\Rightarrow0x=8\left(loai\right)\\x-5=-\left(x+3\right)\Rightarrow x-5=-x-3\Rightarrow x+x=-3+5\Rightarrow2x=2\Rightarrow x=1\left(chon\right)\end{cases}}\)
Vậy x = 1