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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
c: Ta có: \(\sqrt{x-1}+\sqrt{9x-9}-\sqrt{4x-4}=4\)
\(\Leftrightarrow2\sqrt{x-1}=4\)
\(\Leftrightarrow x-1=4\)
hay x=5
e: Ta có: \(\sqrt{4x^2-28x+49}-5=0\)
\(\Leftrightarrow\left|2x-7\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-7=5\\2x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=1\end{matrix}\right.\)
a. ĐKXĐ: $x\in\mathbb{R}$
PT $\Leftrightarrow \sqrt{(x-2)^2}=2-x$
$\Leftrightarrow |x-2|=2-x$
$\Leftrightarrow 2-x\geq 0$
$\Leftrightarrow x\leq 2$
b. ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow \sqrt{4}.\sqrt{x-2}-\frac{1}{5}\sqrt{25}.\sqrt{x-2}=3\sqrt{x-2}-1$
$\Leftrightarrow 2\sqrt{x-2}-\sqrt{x-2}=3\sqrt{x-2}-1$
$\Leftrightarrow 1=2\sqrt{x-2}$
$\Leftrightarrow \frac{1}{2}=\sqrt{x-2}$
$\Leftrightarrow \frac{1}{4}=x-2$
$\Leftrightarrow x=\frac{9}{4}$ (tm)
a) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐK: \(x\ge1\))
\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow-2\sqrt{x-1}=-2\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{2}{2}\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\left(tm\right)\)
b) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\) (ĐK: \(x\ge-1\))
\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
\(\Leftrightarrow x+1=16\)
\(\Leftrightarrow x=15\left(tm\right)\)
a) ĐKXĐ: \(x\ge0\)
Ta có: \(\left(x+3\sqrt{x}+2\right)\left(x+9\sqrt{x}+18\right)=168x\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+6\right)=168x\)
\(\Leftrightarrow\left(x+6\right)^2+12\sqrt{x}\left(x+6\right)-133=0\)
\(\Leftrightarrow\left(x+6\right)^2+19\sqrt{x}\left(x+6\right)-7\sqrt{x}\left(x+6\right)-133=0\)
\(\Leftrightarrow\left(x+6\right)\left(x+19\sqrt{x}+6\right)-7\sqrt{x}\left(x+19\sqrt{x}+6\right)=0\)
\(\Leftrightarrow\left(x-7\sqrt{x}+6\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=36\end{matrix}\right.\)
Dòng thứ 2 qua dòng thứ 3 anh làm chậm lại được không ạ, tại tắt quá e không hiểu
\(a,PT\Leftrightarrow x\sqrt{3}=x+2\\ \Leftrightarrow3x^2=x^2+4x+4\\ \Leftrightarrow2x^2-4x-4=0\Leftrightarrow x^2-2x-2=0\\ \Delta=4+8=12\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2-2\sqrt{3}}{2}=1-\sqrt{3}\\x=\dfrac{2+2\sqrt{3}}{2}=1+\sqrt{3}\end{matrix}\right.\)
\(b,ĐK:x\ge\dfrac{2}{3}\\ PT\Leftrightarrow3x-2=7-4\sqrt{3}\\ \Leftrightarrow3x=9-4\sqrt{3}\\ \Leftrightarrow x=\dfrac{9-4\sqrt{3}}{3}\left(tm\right)\)
\(c,ĐK:x\ge-1\\ PT\Leftrightarrow\left(x+1-4\sqrt{x+1}+4\right)+\left(x^2-6x+9\right)=0\\ \Leftrightarrow\left(\sqrt{x+1}-2\right)^2+\left(x-3\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}\sqrt{x+1}=2\\x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+1=4\\x=3\end{matrix}\right.\Leftrightarrow x=3\left(tm\right)\)
a) ĐK:\(x\ge4\)
\(\sqrt{x-1}+\sqrt{x-4}=\sqrt{x+4}\Leftrightarrow x-1+x-4+2\sqrt{\left(x-1\right)\left(x-4\right)}=x+4\Leftrightarrow9-x=2\sqrt{x^2-5x+4}\left(ĐK:x\le9\right)\Leftrightarrow\left(9-x\right)^2=4\left(x^2-5x+4\right)\Leftrightarrow81-18x+x^2=4x^2-20x+16\Leftrightarrow3x^2-2x-65=0\Leftrightarrow3x^2-15x+13x-65=0\Leftrightarrow3x\left(x-5\right)+13\left(x-5\right)=0\Leftrightarrow\left(x-5\right)\left(3x+13\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x-5=0\\3x+13=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=5\left(tm\right)\\x=-\dfrac{13}{3}\left(ktm\right)\end{matrix}\right.\)
Vậy S={5}
b)\(\sqrt[3]{2x-1}+\sqrt[3]{x-1}=1\Leftrightarrow\sqrt[3]{2x-1}-1+\sqrt[3]{x-1}=0\Leftrightarrow\dfrac{2x-1-1}{\left(\sqrt[3]{2x-1}\right)^2+2.\sqrt[3]{2x-1}+1}+\dfrac{x-1}{\left(\sqrt[3]{x-1}\right)^2}=0\Leftrightarrow\left(x-1\right)\left[\dfrac{2}{\left(\sqrt[3]{2x-1}+2.\sqrt[3]{2x-1}+1\right)}+\dfrac{1}{\left(\sqrt[3]{x-1}\right)^2}\right]=0\)(1)
Dễ thấy \(\dfrac{2}{\left(\sqrt[3]{2x-1}+2.\sqrt[3]{2x-1}+1\right)}+\dfrac{1}{\left(\sqrt[3]{x-1}\right)^2}>0\)
Vậy (1)\(\Leftrightarrow x-1=0\Leftrightarrow x=1\)
Vậy S={1}
c) ĐK:\(\left[{}\begin{matrix}x\le-4\\x\ge-1\end{matrix}\right.\)
\(5\sqrt{x^2+5x+8}=x^2+5x+4\left(2\right)\)
Đặt a=x2+5x+4(a\(\ge0\))
(2)\(\Leftrightarrow5\sqrt{a+4}=a\Leftrightarrow25\left(a+4\right)=a^2\Leftrightarrow a^2-25a-100=0\Leftrightarrow\)\(\left[{}\begin{matrix}a=\dfrac{25+5\sqrt{41}}{2}\left(tm\right)\\a=\dfrac{25-5\sqrt{41}}{2}\left(ktm\right)\end{matrix}\right.\)\(\Leftrightarrow a=\dfrac{25+5\sqrt{41}}{2}\Leftrightarrow\dfrac{25+5\sqrt{41}}{2}=x^2+5x+4\Leftrightarrow25+5\sqrt{41}=2x^2+10x+8\Leftrightarrow2x^2+10x-17-5\sqrt{41}=0\Leftrightarrow\)\(\left[{}\begin{matrix}x=3,045972466\left(tm\right)\\x=-8,045972466\left(tm\right)\end{matrix}\right.\)
Vậy S={-8,045972466;3,045972466}
c) ĐK:\(\left[{}\begin{matrix}x\le-4\\x\ge-1\end{matrix}\right.\)
\(5\sqrt{x^2+5x+28}=x^2+5x+4\left(1\right)\)
Đặt a=x2+5x+4(a\(\ge0\))
Vậy \(\left(1\right)\Leftrightarrow5\sqrt{a+24}=a\Leftrightarrow25\left(a+24\right)=a^2\Leftrightarrow a^2-25a-600=0\Leftrightarrow a^2-40a+15a-600=0\Leftrightarrow a\left(a-40\right)+15\left(a-40\right)=0\Leftrightarrow\left(a-40\right)\left(a+15\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}a-40=0\\a+15=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}a=40\left(tm\right)\\a=-15\left(ktm\right)\end{matrix}\right.\)
Vậy ta có a=40\(\Leftrightarrow x^2+5x+4=40\Leftrightarrow x^2+5x-36=0\Leftrightarrow x^2-4x+9x-36=0\Leftrightarrow x\left(x-4\right)+9\left(x-4\right)=0\Leftrightarrow\left(x-4\right)\left(x+9\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x-4=0\\x+9=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=4\left(tm\right)\\x=-9\left(tm\right)\end{matrix}\right.\)
Vậy S={-9;4}
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...
a. \(\sqrt{x^2+1}=4\)
\(\Leftrightarrow x^2+1=16\)
\(\Leftrightarrow x^2=15\)
\(\Leftrightarrow x=\pm\sqrt{15}\)
b. \(\sqrt{x^2+5x+20}=4\)
\(\Leftrightarrow x^2+5x+20=16\)
\(\Leftrightarrow x^2+5x+4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-4\end{matrix}\right.\)
c. \(x-5\sqrt{x}+4=0\)
Đặt \(t=\sqrt{x}\ge0\)
\(\Rightarrow t^2-5t+4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=4\\t=1\end{matrix}\right.\) (TM)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)
d. \(\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)=8\)
\(\Leftrightarrow x+3\sqrt{x}+\sqrt{x}+3=8\)
\(\Leftrightarrow x+4\sqrt{x}-5=0\)
Tương tự câu c ta được: \(\left[{}\begin{matrix}x=1\left(TM\right)\\x=-1\left(KTM\right)\end{matrix}\right.\)
a) \(\sqrt{x^2+1}=4\)
\(\Leftrightarrow x^2+1=16\)
\(\Leftrightarrow x^2=15\)
\(\Leftrightarrow x=\pm\sqrt{15}\)
b) \(\sqrt{x^2+5x+20}=4\)
\(\Leftrightarrow x^2+5x+20=16\)
\(\Leftrightarrow x^2+5x+4=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-4\end{matrix}\right.\)
c) \(x-5\sqrt{x}+4=0\)(ĐK: \(x\ge0\))
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=16\end{matrix}\right.\)(thỏa mãn)
d) \(\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)=8\)(ĐK: \(x\ge0\))
\(\Leftrightarrow x+4\sqrt{x}-5=0\)
\(\Leftrightarrow\left(\sqrt{x}+5\right)\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow x=1\) (thỏa mãn)