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a: \(\left\{{}\begin{matrix}\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{24}{x-3}-\dfrac{10}{y+2}=126\\\dfrac{24}{x-3}+\dfrac{45}{y+2}=-39\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-55}{y+2}=165\\\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y+2=\dfrac{-1}{3}\\\dfrac{12}{x-3}=48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{7}{3}\\x=\dfrac{13}{4}\end{matrix}\right.\)
Bài 2:
Ta có: \(A=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)
\(=\dfrac{\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}-2}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+1+3-\sqrt{5}-2}{\sqrt{2}}=\sqrt{2}\)
d. \(\sqrt{9x^2+12x+4}=4\)
<=> \(\sqrt{\left(3x+2\right)^2}=4\)
<=> \(|3x+2|=4\)
<=> \(\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)
c: Ta có: \(\dfrac{5\sqrt{x}-2}{8\sqrt{x}+2.5}=\dfrac{2}{7}\)
\(\Leftrightarrow35\sqrt{x}-14=16\sqrt{x}+5\)
\(\Leftrightarrow x=1\)
a) Ta có: \(\left\{{}\begin{matrix}\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\\\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\\\dfrac{5}{x-1}-\dfrac{15}{y-1}=90\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{16}{y-1}=-80\\\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y-1=\dfrac{-1}{5}\\\dfrac{1}{x-1}=18+\dfrac{3}{y-1}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{4}{5}\\x-1=\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=\dfrac{4}{5}\end{matrix}\right.\)
Lời giải:
ĐKXĐ: $x\geq 5$
$2x^2-8x-6=2\sqrt{x-5}\leq (x-5)+1$ theo BĐT Cô-si
$\Leftrightarrow 2x^2-9x-2\leq 0$
$\Leftrightarrow 2x(x-5)+(x-2)\leq 0$
Điều này vô lý do $2x(x-5)\geq 0; x-2\geq 3>0$ với mọi $x\geq 5$
Vậy pt vô nghiệm nên không có đáp án nào đúng.
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)
Bài 1:
a. ĐKXĐ: $x\geq \frac{2}{5}$
PT $\Leftrightarrow 5x-2=7^2=49$
$\Leftrightarrow 5x=51$
$\Leftrightarrow x=\frac{51}{5}=10,2$
b. ĐKXĐ: $x\geq 3$
PT $\Leftrightarrow \sqrt{9(x-3)}+\sqrt{25(x-3)}=24$
$\Leftrightarrow 3\sqrt{x-3}+5\sqrt{x-3}=24$
$\Leftrightarrow 8\sqrt{x-3}=24$
$\Leftrightarrow \sqrt{x-3}=3$
$\Leftrightarrow x-3=9$
$\Leftrightarrow x=12$ (tm)
Bài 1:
c. ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow x^2-5x+6-2(\sqrt{x-2}-1)=0$
$\Leftrightarrow (x-2)(x-3)-2.\frac{x-3}{\sqrt{x-2}+1}=0$
$\Leftrightarrow (x-3)[(x-2)-\frac{2}{\sqrt{x-2}+1}]=0$
$x-3=0$ hoặc $x-2=\frac{2}{\sqrt{x-2}+1}$
Nếu $x-3=0$
$\Leftrightarrow x=3$ (tm)
Nếu $x-2=\frac{2}{\sqrt{x-2}+1}$
$\Leftrightarrow a^2=\frac{2}{a+1}$ (đặt $\sqrt{x-2}=a$)
$\Leftrightarrow a^3+a^2-2=0$
$\Leftrightarrow a^2(a-1)+2a(a-1)+2(a-1)=0$
$\Leftrightarrow (a-1)(a^2+2a+2)=0$
Hiển nhiên $a^2+2a+2=(a+1)^2+1>0$ với mọi $a$ nên $a-1=0$
$\Leftrightarrow a=1\Leftrightarrow \sqrt{x-2}=1\Leftrightarrow x=3$ (tm)
Vậy pt có nghiệm duy nhất $x=3$.
Đặt 1/căn x-3=a; 1/|2y-1|=b
Theo đề, ta có; 8a+11b=5 và 4a+b=3
=>a=7/9; b=-1/9
=>|2y-1|=-9(loại)
=>Hệ vô nghiệm
a, ĐKXĐ: \(x\ge-\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{3}{2}.2\sqrt{1+3x}-\dfrac{5}{3}.3\sqrt{1+3x}-\dfrac{1}{4}.4\sqrt{1+3x}=1\\ \Leftrightarrow3\sqrt{1+3x}-5\sqrt{1+3x}-\sqrt{1+3x}=1\\ \Leftrightarrow-3\sqrt{1+3x}=1\\ \Leftrightarrow\sqrt{1+3x}=-\dfrac{1}{3}\left(vô.lí\right)\)
b, \(\Leftrightarrow\sqrt{\left(x-\dfrac{1}{2}\right)^2}=3\\ \Leftrightarrow\left|x-\dfrac{1}{2}\right|=3\\ \Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=3\\x-\dfrac{1}{2}=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
a) ĐKXĐ: \(x\ge-\dfrac{1}{3}\)
\(pt\Leftrightarrow3\sqrt{3x+1}-5\sqrt{3x+1}-\sqrt{3x+1}=1\)
\(\Leftrightarrow-3\sqrt{3x+1}=1\Leftrightarrow\sqrt{3x+1}=-\dfrac{1}{3}\left(VLý\right)\)
Vậy \(S=\varnothing\)
b) \(pt\Leftrightarrow\sqrt{\left(x-\dfrac{1}{2}\right)^2}=3\Leftrightarrow\left|x-\dfrac{1}{2}\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=3\\x-\dfrac{1}{2}=-3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
ĐK: `x>=0 ; x \ne 25/49`
`(3\sqrtx+1)/(7\sqrtx-5)=8/15`
`<=>15(3\sqrtx+1)=8(7\sqrtx-5)`
`<=>45\sqrtx+15=56\sqrtx-40`
`<=>11\sqrtx=55`
`<=>\sqrtx=5`
`<=>x=25`
Vậy `S={25}`.
Ta có: \(\dfrac{3\sqrt{x}+1}{7\sqrt{x}-5}=\dfrac{8}{15}\)
\(\Leftrightarrow56\sqrt{x}-40-45\sqrt{x}-15=0\)
\(\Leftrightarrow11\sqrt{x}=55\)
hay x=25