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Lời giải:
a) ĐK: $x\geq 2$
PT $\Leftrightarrow \sqrt{(x-2)(x+2)}-3\sqrt{x-2}=0$
$\Leftrightarrow \sqrt{x-2}(\sqrt{x+2}-3)=0$
\(\Rightarrow \left[\begin{matrix} \sqrt{x-2}=0\\ \sqrt{x+2}-3=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=2\\ x=7\end{matrix}\right.\) (thỏa mãn)
Vậy..........
b) ĐK: $x\geq 0$
PT $\Leftrightarrow (\sqrt{x}-3)^2=0$
$\Leftrightarrow \sqrt{x}-3=0$
$\Leftrightarrow x=9$ (thỏa mãn)
c) ĐK: $x\geq 3$
PT $\Leftrightarrow \sqrt{9(x-3)}+\sqrt{x-3}-\frac{1}{2}\sqrt{4(x-3)}=7$
$\Leftrightarrow 3\sqrt{x-3}+\sqrt{x-3}-\sqrt{x-3}=7$
$\Leftrightarrow 3\sqrt{x-3}=7$
$\Leftrightarrow x-3=(\frac{7}{3})^2$
$\Rightarrow x=\frac{76}{9}$
d)
ĐK: $x\geq \frac{-1}{2}$
PT $\Leftrightarrow 3\sqrt{4(2x+1)}-\frac{1}{3}\sqrt{9(2x+1)}-\frac{1}{2}\sqrt{25(2x+1)}+\sqrt{\frac{1}{4}(2x+1)}=6$
$\Leftrightarrow 6\sqrt{2x+1}-\sqrt{2x+1}-\frac{5}{2}\sqrt{2x+1}+\frac{1}{2}\sqrt{2x+1}=6$
$\Leftrightarrow 3\sqrt{2x+1}=6$
$\Leftrightarrow \sqrt{2x+1}=2$
$\Rightarrow x=\frac{3}{2}$ (thỏa mãn)
a, ĐK :a >= 3
\(25\sqrt{\frac{a-3}{25}}-7\sqrt{\frac{4a-12}{9}}-7\sqrt{a^2-9}+18\sqrt{\frac{9a^2-81}{81}}=0\)
\(\Leftrightarrow5\sqrt{a-3}-\frac{14}{3}\sqrt{a-3}-7\sqrt{\left(a-3\right)\left(a+3\right)}+6\sqrt{\left(a-3\right)\left(a+3\right)}=0\)
\(\Leftrightarrow\sqrt{a-3}\left(5-\frac{14}{3}-\sqrt{a+3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{a-3}=0\\\sqrt{a+3}=\frac{1}{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}a=3\left(tm\right)\\a=-\frac{2}{9}\left(loai\right)\end{cases}}\)
b, \(ĐK:x\ge-\frac{1}{2}\)
\(\Leftrightarrow3\sqrt{2x+1}-2\sqrt{2x+1}+\frac{1}{3}\sqrt{2x+1}=4\)
\(\Leftrightarrow\frac{4}{3}\sqrt{2x+1}=4\)
\(\Leftrightarrow\sqrt{2x+1}=3\)
\(\Leftrightarrow x=4\left(tm\right)\)
a) đk: \(a\ge3\)
pt \(\Leftrightarrow25\frac{\sqrt{a-3}}{\sqrt{25}}-7\frac{\sqrt{4\left(a-3\right)}}{\sqrt{9}}-7\sqrt{a^2-9}+18\frac{\sqrt{9\left(a^2-9\right)}}{\sqrt{81}}=0\)
\(\Leftrightarrow5\sqrt{a-3}-\frac{7.2}{3}\sqrt{a-3}-7\sqrt{a^2-9}+\frac{18.3}{9}\sqrt{a^2-9}=0\)
\(\Leftrightarrow5\sqrt{a-3}-\frac{14}{3}\sqrt{a-3}-7\sqrt{a^2-9}+6\sqrt{a^2-9}=0\)
\(\Leftrightarrow\frac{1}{3}\sqrt{a-3}-\sqrt{a^2-9}=0\)
\(\Leftrightarrow\frac{1}{3}\sqrt{a-3}=\sqrt{a^2-9}\)
\(\Leftrightarrow\frac{1}{9}\left(a-3\right)=a^2-9\)
\(\Leftrightarrow a^2-\frac{1}{9}a-\frac{26}{3}=0\Leftrightarrow\orbr{\begin{cases}a=3\left(tm\right)\\a=-\frac{26}{9}\left(loại\right)\end{cases}}\)
a) ĐKXĐ : \(x\ge0\)
Ta có : \(\sqrt{3x}-\sqrt{27}+\sqrt{75x}=3\Leftrightarrow\sqrt{x}\left(\sqrt{3}+\sqrt{75}\right)=3+\sqrt{27}\)
\(\Leftrightarrow\sqrt{x}=\frac{3+\sqrt{27}}{\sqrt{3}+\sqrt{75}}=\frac{\sqrt{3}+3}{6}\)
\(\Leftrightarrow x=\frac{\left(3+\sqrt{3}\right)^2}{36}\)
b) ĐKXĐ : \(x\ge1\)
\(\sqrt{x-1}-\sqrt{4x-4}+\sqrt{9x-9}=10\)
\(\Leftrightarrow\sqrt{x-1}-\sqrt{4.\left(x-1\right)}+\sqrt{9.\left(x-1\right)}=10\)
\(\Leftrightarrow\sqrt{x-1}-2\sqrt{x-1}+3\sqrt{x-1}=10\)
\(\Leftrightarrow\sqrt{x-1}=5\Leftrightarrow x=26\) (TMĐK)
c) ĐKXĐ: \(x\ge-\frac{1}{2}\)
\(\sqrt{2x+1}+\sqrt{18x+9}-\sqrt{50x+25}=-3\)
\(\Leftrightarrow\sqrt{2x+1}+\sqrt{9\left(2x+1\right)}-\sqrt{25\left(2x+1\right)}=-3\)
\(\Leftrightarrow\sqrt{2x+1}+3\sqrt{2x+1}-5\sqrt{2x+1}=-3\)
\(\Leftrightarrow0=-3\) (Vô lí - loại)
Vậy pt vô nghiệm.
\(\sqrt{x-1}=5\)
\(\Leftrightarrow x-1=25\) (bình phương 2 vế)
\(\Leftrightarrow x=26\)
Lời giải:
a) ĐK: \(x>0; x\neq 25; x\neq 36\)
PT \(\Rightarrow (\sqrt{x}-2)(\sqrt{x}-6)=(\sqrt{x}-5)(\sqrt{x}-4)\)
\(\Leftrightarrow x-8\sqrt{x}+12=x-9\sqrt{x}+20\)
\(\Leftrightarrow \sqrt{x}=8\Rightarrow x=64\) (thỏa mãn)
Vậy.......
b)
ĐK: \(x\geq \frac{-1}{2}\)
PT \(\Leftrightarrow \sqrt{9(2x+1)}-\sqrt{4(2x+1)}+\frac{1}{3}\sqrt{2x+1}=4\)
\(\Leftrightarrow 3\sqrt{2x+1}-2\sqrt{2x+1}+\frac{1}{3}\sqrt{2x+1}=4\)
\(\Leftrightarrow \frac{4}{3}\sqrt{2x+1}=4\Leftrightarrow \sqrt{2x+1}=3\)
\(\Rightarrow x=\frac{3^2-1}{2}=4\) (thỏa mãn)
c)
ĐK: \(x\geq 2\)
PT \(\Leftrightarrow \sqrt{4(x-2)}-\frac{1}{2}\sqrt{x-2}+\sqrt{9(x-2)}=9\)
\(\Leftrightarrow 2\sqrt{x-2}-\frac{1}{2}\sqrt{x-2}+3\sqrt{x-2}=9\)
\(\Leftrightarrow \frac{9}{2}\sqrt{x-2}=9\Leftrightarrow \sqrt{x-2}=2\Rightarrow x=2^2+2=6\) (thỏa mãn)
Bài 2:Giải phương trình
a,\(\sqrt{8x-4}-2\sqrt{18x-9}+2\sqrt{32x-16}=12\)
b.\(\sqrt{x^2-6x+9}=2x-1\)
phần a đây nhé \(a,\sqrt{4\left(2x-1\right)}-2\sqrt{9\left(2x-1\right)}+2\sqrt{16\left(2x-1\right)}=12\Leftrightarrow2\sqrt{2x-1}-6\sqrt{2x-1}+8\sqrt{2x-1}=12\Leftrightarrow4\sqrt{2x-1}=12\Leftrightarrow\sqrt{2x-1}=3\Leftrightarrow\left\{{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
ĐKXĐ:
a/ \(x-2020>0\Rightarrow x>2020\)
b/ \(x\ne0\)
c/ \(3x+5< 0\Rightarrow x< -\frac{5}{3}\)
d/ \(\frac{x-3}{1-x}\ge0\Rightarrow1< x\le3\)
Bài 2: ĐKXĐ tự tìm
a/ \(2\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=28\)
\(\Leftrightarrow13\sqrt{2x}=28\Rightarrow\sqrt{2x}=\frac{28}{13}\)
\(\Rightarrow x=\frac{392}{169}\)
b/ \(2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}=2\Rightarrow x=9\)
c/ \(3\sqrt{2x+1}>15\Rightarrow\sqrt{2x+1}>5\)
\(\Rightarrow2x+1>25\Rightarrow x>12\)
d/ \(\sqrt{x}+1>12\Rightarrow\sqrt{x}>11\Rightarrow x>121\)
1. a) Ta có: \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}=\frac{x+3}{2}\)
\(\Leftrightarrow\sqrt{x+2\sqrt{x-1}}.2=\frac{x+3}{2}\)
\(\Leftrightarrow\sqrt{x+\sqrt{2-1}}.2=\frac{x+3}{2}\)
Bạn tự khai triển ra nha!
b) Tương tự
2) Tự làm
Ps: Ms lớp 6 nên chỉ làm được như vậy thôi! Bạn tự khai triển thành bài nhé!
1)
a) đk x>=1
\(\Leftrightarrow\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}=\frac{x+3}{2}\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=\frac{x+3}{2}\)
\(\Leftrightarrow\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|=\frac{x+3}{2}\)
\(\Leftrightarrow\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|=\frac{x+3}{2}\)
vs x>=2
thì pt có dạng
\(\sqrt{x-1}+1+\sqrt{x-1}-1=\frac{x+3}{2}\)
\(4\sqrt{x-1}=x+3\)
\(16x-16=x^2+6x+9\)
\(x^2-10x+25=0\)
x=5(tm)
vs 0<=x<1
pt \(2=\frac{x+3}{2}\)
\(x+3=4\)
\(x=1\)
Điều kiện: \(2x-3\ge0\Leftrightarrow x\ge\dfrac{3}{2}\)
\(3\sqrt{2x-3}+2\sqrt{8x-12}=\sqrt{18x-27}+9\)
\(3\sqrt{2x-3}+2\sqrt{4\left(2x-3\right)}-\sqrt{9\left(2x-3\right)}=9\)
\(3\sqrt{2x-3}+4\sqrt{2x-3}-3\sqrt{2x-3}=9\)
\(4\sqrt{2x-3}=9\)
\(x\ge\dfrac{3}{2}\)\(\Rightarrow16\left(2x-3\right)=81\)
\(2x-3=\dfrac{81}{16}\Leftrightarrow x=\dfrac{\dfrac{81}{16}+3}{2}=\dfrac{129}{32}\)