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a)\(\sqrt{4x+20}\) +\(\sqrt{x-5}\) -\(\dfrac{1}{3}\)\(\sqrt{9x-45}\)=4 ; ĐKXĐ : x ≥_+ 5
⇔ \(\sqrt{2^2x+2^2.5}\) +\(\sqrt{x-5}\) -\(\dfrac{1}{3}\)\(\sqrt{3^2x-3^2.5}\) =4
⇔ 2\(\sqrt{x+5}\) +\(\sqrt{x-5}\) -\(\dfrac{1}{3}\)3\(\sqrt{x-5}\) =4 ⇔ 2\(\sqrt{x+5}\) +\(\sqrt{x-5}\) -\(\sqrt{x-5}\) =4⇔2\(\sqrt{x+5}\)=4(tm)
⇔\(\sqrt{x+5}\)=2⇔x+5=4 ⇔x=-1
Vậy x=-1
b) \(\sqrt{x^2-36}\) - \(\sqrt{x-6}\) =0 ; ĐKXĐ: x≥_+6
⇔ \(\sqrt{\left(x-6\right)\left(x+6\right)}\) - \(\sqrt{x-6}\) =0 ⇔ \(\sqrt{x-6}\).\(\sqrt{x+6}\) - \(\sqrt{x-6}\) =0
⇔ \(\sqrt{x-6}\)(\(\sqrt{x+6}\) -1 )=0 ⇔\([\) \(\begin{matrix}\sqrt{x-6}&=0\\\sqrt{x+6}-1&=0\end{matrix}\) ⇔ \([\) \(\begin{matrix}x-6&=0\\x+6-1&=0\end{matrix}\) ⇔\([\) \(\begin{matrix}x&=6\left(ktm\right)\\x&=-5\left(tm\right)\end{matrix}\)
Vậy x=-5
c) \(\sqrt{4-x^2}\) -x +2 =0 ; ĐKXĐ: -2≤x≤2
⇔ \(\sqrt{\left(2-x\right)\left(2+x\right)}\) -x+2 =0 ⇔ \(\sqrt{\left(2-x\right)\left(2+x\right)}\) -(x-2)=0
⇔ \(\sqrt{\left(2-x\right)\left(2+x\right)}\) =(x-2) ⇔ (2-x)(2+x)=(x-2)2 ⇔ 4-x2 = x2-4x+4 ⇔ -x2-x2+4x=4-4
⇔-2x2+4x=0 ⇔ -2x(x-2)=0 ⇔ \([\) \(\begin{matrix}-2x&=0\\x-2&=0\end{matrix}\) ⇔\([\) \(\begin{matrix}x&=0\left(tm\right)\\x&=2\left(tm\right)\end{matrix}\)
Vậy S=\(\left\{0;2\right\}\)
d) \(\sqrt{\left(2x-3\right)\left(x-1\right)}-\sqrt{x-1}=0\) ; ĐKXĐ: x≥\(\dfrac{3}{2}\);x ≥ 1
⇔\(\sqrt{2x-3}.\sqrt{x-1}-\sqrt{x-1}=0\) ⇔ \(\sqrt{x-1}.\left(\sqrt{2x-3}-1\right)=0\)
⇔ \(\left[{}\begin{matrix}\sqrt{x-1}=0\\\sqrt{2x-3}-1=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x-1=0\\2x-3-1=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=1\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
Vậy s=\(\left\{1:2\right\}\)
\(a,\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)\(ĐKXĐ:x\ge-\frac{5}{7}\)
\(\Leftrightarrow9x-7=7x+5\)
\(\Leftrightarrow9x-7x=5+7\)
\(\Leftrightarrow2x=12\)
\(\Leftrightarrow x=6\)
\(b,\sqrt{4x-20}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=4\)
\(\Leftrightarrow\sqrt{4\left(x-5\right)}+3.\frac{\sqrt{x-5}}{\sqrt{9}}-\frac{1}{3}\sqrt{9\left(x-5\right)}=4\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}\left(2+1-1\right)=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\)
\(\Leftrightarrow x=9\)
a) \(\frac{3}{4}\sqrt{x}-\sqrt{9x}+5=\frac{1}{4}\sqrt{9x}\)
ĐK : x ≥ 0
⇔ \(\frac{3}{4}\sqrt{x}-\sqrt{3^2x}-\frac{1}{4}\sqrt{3^2x}=-5\)
⇔ \(\frac{3}{4}\sqrt{x}-3\sqrt{x}-\frac{1}{4}\cdot3\sqrt{x}=-5\)
⇔ \(-\frac{9}{4}\sqrt{x}-\frac{3}{4}\sqrt{x}=-5\)
⇔ \(-3\sqrt{x}=-5\)
⇔ \(\sqrt{x}=15\)
⇔ \(x=225\)( tm )
b) \(\sqrt{3-x}-\sqrt{27-9x}+1,25\sqrt{48-16x}=6\)
ĐK : x ≤ 3
⇔ \(\sqrt{3-x}-\sqrt{3^2\left(3-x\right)}+\frac{5}{4}\sqrt{4^2\left(3-x\right)}=6\)
⇔ \(\sqrt{3-x}-3\sqrt{3-x}+\frac{5}{4}\cdot4\sqrt{3-x}=6\)
⇔ \(-2\sqrt{3-x}+5\sqrt{3-x}=6\)
⇔ \(3\sqrt{3-x}=6\)
⇔ \(\sqrt{3-x}=2\)
⇔ \(3-x=4\)
⇔ \(x=-1\)( tm )
c) \(\sqrt{9x^2+12x+4}=4\)
⇔ \(\sqrt{\left(3x+2\right)^2}=4\)
⇔ \(\left|3x+2\right|=4\)
⇔ \(\orbr{\begin{cases}3x+2=4\\3x+2=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-2\end{cases}}\)
d) \(\frac{1}{3}\sqrt{x-1}+2\sqrt{4x-4}-12\sqrt{\frac{x-1}{25}}=\frac{29}{15}\)
ĐK : x ≥ 1
⇔ \(\frac{1}{3}\sqrt{x-1}+2\sqrt{2^2\left(x-1\right)}-12\sqrt{\left(\frac{1}{5}\right)^2\cdot\left(x-1\right)}=\frac{29}{15}\)
⇔ \(\frac{1}{3}\sqrt{x-1}+2\cdot2\sqrt{x-1}-12\cdot\frac{1}{5}\sqrt{x-1}=\frac{29}{15}\)
⇔ \(\frac{1}{3}\sqrt{x-1}+4\sqrt{x-1}-\frac{12}{5}\sqrt{x-1}=\frac{29}{15}\)
⇔ \(\frac{29}{15}\sqrt{x-1}=\frac{29}{15}\)
⇔ \(\sqrt{x-1}=1\)
⇔ \(x-1=1\)
⇔ \(x=2\)( tm )
a) ĐK: \(x\ge5\)
\(\sqrt{4x-20}+\frac{1}{3}\sqrt{9x-45}-\frac{1}{5}\sqrt{16x-80}=0\)
\(\Leftrightarrow\)\(\sqrt{4\left(x-5\right)}+\frac{1}{3}\sqrt{9\left(x-5\right)}-\frac{1}{5}\sqrt{16\left(x-5\right)}=0\)
\(\Leftrightarrow\)\(2\sqrt{x-5}+\sqrt{x-5}-\frac{4}{5}\sqrt{x-5}=0\)
\(\Leftrightarrow\)\(\frac{11}{5}\sqrt{x-5}=0\)
\(\Leftrightarrow\)\(x-5=0\)
\(\Leftrightarrow\)\(x=5\) (t/m)
Vậy
b) \(-5x+7\sqrt{x}=-12\)
\(\Leftrightarrow\)\(5x-7\sqrt{x}-12=0\)
\(\Leftrightarrow\)\(\left(\sqrt{x}+1\right)\left(5\sqrt{x}-12\right)=0\)
đến đây tự làm
c) d) e) bạn bình phương lên
f) \(VT=\sqrt{3\left(x^2+2x+1\right)+9}+\sqrt{5\left(x^4-2x^2+1\right)+25}\)
\(=\sqrt{3\left(x+1\right)^2+9}+\sqrt{5\left(x^2-1\right)^2}\)
\(\ge\sqrt{9}+\sqrt{25}=8\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}x+1=0\\x^2-1=0\end{cases}}\)\(\Leftrightarrow\)\(x=-1\)
Vậy...
Dài Vãi mik ko bít giải phhương trình sorry nha
a) \(\sqrt{\left(2x-1\right)^2}=3\)
⇔ \(\left|2x-1\right|=3\)
⇔ \(\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\)
⇔ \(\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)
b) \(3\sqrt{x}-2\sqrt{9x}+\sqrt{16x}=5\)
ĐKXĐ : \(x\ge0\)
⇔ \(3\sqrt{x}-2\sqrt{3^2x}+\sqrt{4^2x}=5\)
⇔ \(3\sqrt{x}-2\cdot3\sqrt{x}+4\sqrt{x}=5\)
⇔ \(7\sqrt{x}-6\sqrt{x}=5\)
⇔ \(\sqrt{x}=5\)
⇔ \(x=25\)( tm )
c) \(\sqrt{4x+20}-3\sqrt{5+x}+\frac{3}{4}\sqrt{9x+45}=6\)
ĐKXĐ : \(x\ge-5\)
⇔ \(\sqrt{2^2\left(x+5\right)}-3\sqrt{x+5}+\frac{3}{4}\sqrt{3^2\left(x+5\right)}=6\)
⇔ \(2\sqrt{x+5}-3\sqrt{x+5}+\frac{3}{4}\cdot3\sqrt{x+5}=6\)
⇔ \(-\sqrt{x+5}+\frac{9}{4}\sqrt{x+5}=6\)
⇔ \(\frac{5}{4}\sqrt{x+5}=6\)
⇔ \(\sqrt{x+5}=\frac{24}{5}\)
⇔ \(x+5=\frac{576}{25}\)
⇔ \(x=\frac{451}{25}\left(tm\right)\)