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22 tháng 7 2019
\(ĐKXĐ:x;y\ge2\)
\(\hept{\begin{cases}\sqrt{x-2}-y\sqrt{y}=\sqrt{y-2}-x\sqrt{x}\left(1\right)\\3x^2-y^2-xy-7x+y+5=0\left(2\right)\end{cases}}\)
Giải \(\left(1\right)\Leftrightarrow\sqrt{x-2}-\sqrt{y-2}+x\sqrt{x}-y\sqrt{y}=0\)
\(\Leftrightarrow\frac{x-2-y+2}{\sqrt{x-2}+\sqrt{y-2}}+\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)=0\)
\(\Leftrightarrow\frac{x-y}{\sqrt{x-2}+\sqrt{y-2}}+\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-\sqrt{y}\right)\left(\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x-2}+\sqrt{y-2}}+x+\sqrt{xy}+y\right)=0\)
Kết hợp ĐKXĐ dễ thấy cái ngoặc to luôn dương
Nên \(\sqrt{x}-\sqrt{y}=0\Rightarrow x=y\)
Thay vào pt (2) đc
\(3x^2-x^2-x^2-7x+x+5=0\)
\(\Leftrightarrow x^2-6x+5=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\Rightarrow y=1\left(thoa\cdot man\cdot DKXD\right)\\x=5\Rightarrow y=5\left(Thoa\cdot man\cdot DKXD\right)\end{cases}}\)
TQ
20 tháng 2 2018
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7 tháng 2 2018
\(\left(1\right)\Leftrightarrow y=\frac{4x^2}{y+45}\)
\(\Rightarrow\left(2\right)\Leftrightarrow\left(\frac{4x^2}{x+45}\right)^2+95.\frac{4x^2}{x+45}+6-7x^2-5x=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-9\right)\left(3x^2-43x+90\right)=0\)
23 tháng 3 2020
\(\left\{{}\begin{matrix}7x-3y=5\\\frac{x}{2}+\frac{y}{3}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{3y+5}{7}\\\frac{3y+5}{14}+\frac{y}{3}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
Vậy...
AH
Akai Haruma
Giáo viên
16 tháng 5 2018
Lời giải:
Ta có:
\(\left\{\begin{matrix} 2xy+y+2=-8x\\ x^2y^2+xy+1=7x^2\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} 2(xy+1)=-(8x+y)\\ (xy+1)^2=7x^2+xy\end{matrix}\right.\)
\(\Rightarrow \left[\frac{-(8x+y)}{2}\right]^2=7x^2+xy\)
\(\Leftrightarrow 64x^2+y^2+16xy=28x^2+4xy\)
\(\Leftrightarrow 36x^2+y^2+12xy=0\)
\(\Leftrightarrow (6x+y)^2=0\Rightarrow y=-6x\)
Thay vào pt đầu tiên suy ra:
\(-6x^2+x+1=0\Rightarrow \left[\begin{matrix} x=\frac{1}{2}\rightarrow y=-3\\ x=\frac{-1}{3}\Rightarrow y=2\end{matrix}\right.\)
Vậy...........
AH
Akai Haruma
Giáo viên
23 tháng 9 2018
a)
ĐKXĐ: \(x> \frac{-5}{7}\)
Ta có: \(\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)
\(\Rightarrow 9x-7=\sqrt{7x+5}.\sqrt{7x+5}=7x+5\)
\(\Rightarrow 2x=12\Rightarrow x=6\) (hoàn toàn thỏa mãn)
Vậy......
b) ĐKXĐ: \(x\geq 5\)
\(\sqrt{4x-20}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=4\)
\(\Leftrightarrow \sqrt{4}.\sqrt{x-5}+3\sqrt{\frac{1}{9}}.\sqrt{x-5}-\frac{1}{3}\sqrt{9}.\sqrt{x-5}=4\)
\(\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow 2\sqrt{x-5}=4\Rightarrow \sqrt{x-5}=2\Rightarrow x-5=2^2=4\Rightarrow x=9\)
(hoàn toàn thỏa mãn)
Vậy..........
AH
Akai Haruma
Giáo viên
23 tháng 9 2018
c) ĐK: \(x\in \mathbb{R}\)
Đặt \(\sqrt{6x^2-12x+7}=a(a\geq 0)\Rightarrow 6x^2-12x+7=a^2\)
\(\Rightarrow 6(x^2-2x)=a^2-7\Rightarrow x^2-2x=\frac{a^2-7}{6}\)
Khi đó:
\(2x-x^2+\sqrt{6x^2-12x+7}=0\)
\(\Leftrightarrow \frac{7-a^2}{6}+a=0\)
\(\Leftrightarrow 7-a^2+6a=0\)
\(\Leftrightarrow -a(a+1)+7(a+1)=0\Leftrightarrow (a+1)(7-a)=0\)
\(\Rightarrow \left[\begin{matrix} a=-1\\ a=7\end{matrix}\right.\) \(\Rightarrow a=7\) vì \(a\geq 0\)
\(\Rightarrow 6x^2-12x+7=a^2=49\)
\(\Rightarrow 6x^2-12x-42=0\Leftrightarrow x^2-2x-7=0\)
\(\Leftrightarrow (x-1)^2=8\Rightarrow x=1\pm 2\sqrt{2}\)
(đều thỏa mãn)
Vậy..........
S
1
MT
25 tháng 12 2015
ĐK: \(0\le x\le7\)
\(2\sqrt{3}-\sqrt{7x-x^2}=0\Leftrightarrow2\sqrt{3}=\sqrt{7x-x^2}\Leftrightarrow12=7x-x^2\)
\(\Leftrightarrow x^2-7x+12=0\Leftrightarrow x^2-4x-3x+12=0\Leftrightarrow x.\left(x-4\right)-3.\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\Leftrightarrow x=3\text{ hoặc }x=4\text{. Vậy S=}\left\{3;4\right\}\)
HF
1
AH
Akai Haruma
Giáo viên
31 tháng 10 2020
Lời giải:
ĐK: $x\geq \frac{-18}{7}$
PT $\Leftrightarrow x^2+3x-4-3(\sqrt{x+3}-2)-(\sqrt{7x+18}-5)=0$
$\Leftrightarrow (x-1)(x+4)-3.\frac{x-1}{\sqrt{x+3}+2}-\frac{7(x-1)}{\sqrt{7x+18}+5}=0$
$\Leftrightarrow (x-1)\left(x+4-\frac{3}{\sqrt{x+3}+2}-\frac{7}{\sqrt{7x+18}+5}\right)=0$
Xét các TH:
Nếu $x-1=0\Rightarrow x=1$ (thỏa mãn)
Nếu $x+4-\frac{3}{\sqrt{x+3}+2}-\frac{7}{\sqrt{7x+18}+5}=0$
$\Leftrightarrow (x+2)+1-\frac{3}{\sqrt{x+3}+2}+1-\frac{7}{\sqrt{7x+18}+5}=0$
$\Leftrightarrow x+2+\frac{\sqrt{x+3}-1}{\sqrt{x+3}+2}+\frac{\sqrt{7x+18}-2}{\sqrt{7x+18}+5}=0$
\(\Leftrightarrow (x+2)+\frac{x+2}{(\sqrt{x+3}+1)(\sqrt{x+3}+2)}+\frac{7(x+2)}{(\sqrt{7x+18}+2)(\sqrt{7x+18}+5)}=0\)
\(\Leftrightarrow (x+2)\left( 1+\frac{1}{(\sqrt{x+3}+1)(\sqrt{x+3}+2)}+\frac{7}{(\sqrt{7x+18}+2)(\sqrt{7x+18}+5)}\right)=0\)
Dễ thấy biểu thức trong ngoặc lớn luôn dương nên $x+2=0\Leftrightarrow x=-2$
Vậy $x=-2$ hoặc $x=1$
7 tháng 3 2016
Bài 2 giải như sau (sau khi tác giả đã sửa): Điều kiện \(x,y>0.\)
Từ hệ ta suy ra \(1+\frac{3}{x+3y}=\frac{2}{\sqrt{x}},1-\frac{3}{x+3y}=\frac{4\sqrt{2}}{\sqrt{7y}}.\) Cộng và trừ hai phương trình, chia cả hai vế cho 2, ta sẽ được 2 phương trình \(1=\frac{1}{\sqrt{x}}+\frac{2\sqrt{2}}{\sqrt{7y}},\frac{3}{x+3y}=\frac{1}{\sqrt{x}}-\frac{2\sqrt{2}}{\sqrt{7y}}.\) Nhân hai phương trình với nhau, vế theo vế, ta được
\(\frac{3}{x+3y}=\frac{1}{x}-\frac{8}{7y}\to21xy=\left(x+3y\right)\left(7y-8x\right)\to21y^2-38xy-8x^2=0\to x=\frac{y}{2},x=-\frac{21}{4}y.\)
Đến đây ta được y=2x (trường hợp kia loại). Từ đó thế vào ta được \(1+\frac{3}{7x}=\frac{2}{\sqrt{x}}\to7x-14\sqrt{x}+3=0\to\sqrt{x}=\frac{7\pm2\sqrt{7}}{2}\to...\)
x²-7x+6=0
x.x-7.x=-6
x(x-7)=-6
Vì x€ Z nên x-7€ Z
Mà x( x-7)=-6
=>-6 chia hết cho x
Hay x là ước của -6
=>x€{1,2,3,6,-1,-2,-3,-6}
Vậy x€{1,2,3,6,-1,-2,-3,-6}
\(x^2-7x+6=0\\ x^2-x-6x+6=0\\ \left(x-6\right)\left(x-1\right)=0\\ =>\left[{}\begin{matrix}x=6\\x=1\end{matrix}\right.\)