Giải hệ phưng trình \(\left\{{}\begin{matrix}\sqrt{x-1}+2\sqrt{y}=5\\\frac{4x-9y-4}{2\sqrt{x-1}+3\sqrt{y}}=-4\end{matrix}\right.\)
Giải hệ phương trình:
1. \(\left\{{}\begin{matrix}3\sqrt{x}-\sqrt{y}=5\\2\sqrt{x}+3\sqrt{y}=18\end{matrix}\right.\)
2. \(\left\{{}\begin{matrix}\sqrt{x+3}-2\sqrt{y+1}=2\\2\sqrt{x+3}+\sqrt{y+1}=4\end{matrix}\right.\)
3. \(\left\{{}\begin{matrix}3\sqrt{x}+2\sqrt{y}=6\\\sqrt{x}-\sqrt{y}=4,5\end{matrix}\right.\)
4. \(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y+1}=1\\\sqrt{y}+\sqrt{x+1}=1\end{matrix}\right.\)
1) Ta có: \(\left\{{}\begin{matrix}3\sqrt{x}-\sqrt{y}=5\\2\sqrt{x}+3\sqrt{y}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}9\sqrt{x}-3\sqrt{y}=15\\2\sqrt{x}+3\sqrt{y}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}11\sqrt{x}=33\\3\sqrt{x}-\sqrt{y}=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=3\\\sqrt{y}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=9\\y=16\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=9\\y=16\end{matrix}\right.\)
2) Ta có: \(\left\{{}\begin{matrix}\sqrt{x+3}-2\sqrt{y+1}=2\\2\sqrt{x+3}+\sqrt{y+1}=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-2\sqrt{x+3}+4\sqrt{y+1}=-4\\2\sqrt{x+3}+\sqrt{y+1}=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5\sqrt{y+1}=0\\\sqrt{x+3}-2\sqrt{y+1}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{y+1}=0\\\sqrt{x+3}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y+1=0\\x+3=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=1\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
4. Đk: \(x,y\ge0\)
\(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y+1}=1\\\sqrt{y}+\sqrt{x+1}=1\end{matrix}\right.\left(1\right)\)
Ta có: \(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y+1}\ge0+1=1\\\sqrt{y}+\sqrt{x+1}\ge0+1=1\end{matrix}\right.\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow\left\{{}\begin{matrix}\sqrt{x}=0,\sqrt{x+1}=1\\\sqrt{y}=0,\sqrt{y+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)<tmđk>
Vậy hệ pt có nghiệm \(\left(x,y\right)=\left(0;0\right)\)
giúp mik giải bài hệ pt vs ạ!
1,\(\left\{{}\begin{matrix}x^2+y^2+\dfrac{2xy}{x+y}=1\\\sqrt{x+y}=x^2-y\end{matrix}\right.\)
2,\(\left\{{}\begin{matrix}2x^3+xy^2+x=y^3+4x^2y+2y\\\sqrt{4x^2+x+6}-5\sqrt{1+2y}=1-4y\end{matrix}\right.\)
3,\(\left\{{}\begin{matrix}2x^2+\sqrt{2}x=\left(x+y\right)y+\sqrt{x+y}\\\sqrt{x-1}+xy=\sqrt{y^2+21}\end{matrix}\right.\)
4,\(\left\{{}\begin{matrix}\sqrt{9y^2+\left(2y+3\right)\left(y-x\right)}+4\sqrt{xy}=7x\\\left(2y-1\right)\sqrt{1+x}+\left(2y+1\right)\sqrt{1-x}=2y\end{matrix}\right.\)
1)Điều kiện: \(x + y > 0\)\((1) \Leftrightarrow (x + y)^2 - 2xy + \dfrac{2xy}{x + y} - 1 = 0 \\ \Leftrightarrow (x + y)^3 - 2xy(x + y) + 2xy -(x + y) = 0 \\ \Leftrightarrow (x+y)[(x+y)^2- 1]-2xy(x+y-1)=0 \\ \Leftrightarrow (x+y)(x+y+1)(x+y-1)-2xy(x+y-1)=0 \\ \Leftrightarrow (x + y - 1)[(x+y)(x + y + 1)-2xy] = 0 \\ \Leftrightarrow \left[ \begin{matrix}x + y = 1 \,\, (3) \\ x^2+y^2+x+y=0 \,\, (4) \end{matrix} \right.\)(4) vô nghiệm vì x + y > 0
Thế (3) vào (2) , giải được nghiệm của hệ :\((x =1 ; y = 0)\)và \((x = -2 ; y = 3)\)
\((1)\Leftrightarrow (x-2y)+(2x^3-4x^2y)+(xy^2-2y^3)=0\)\(\Leftrightarrow (x-2y)(1+2x^2+y^2)=0\)
\(\Leftrightarrow x=2y\)(vì \(1+2x^2+y^2>0, \forall x,y\))
Thay vào phương trình (2) giải dễ dàng.
Điều kiện:\(9y^2+(2y+3)(y-x)\geq 0;xy\geq 0;-1\leq x\leq 1\)
Từ phương trình thứ nhất có \(x\geq 0\Rightarrow y\geq 0\)
Xét \(\left\{\begin{matrix} x=0\\ y=0 \end{matrix}\right.\) thỏa mãn hệ
Xét x,y không đồng thời bằng 0, ta có
\(\sqrt{9y^2+(2y+3)(y-x)}-3x+4\sqrt{xy}-4x=0\)
\(\Leftrightarrow \frac{9y^2+(2y+3)(y-x)-9x^2}{\sqrt{9y^2+(2y-3)(y-x)+3x}}+\frac{4(xy-x^2)}{\sqrt{xy}+x}=0\)
\(\Leftrightarrow (y-x)\left [ \frac{11y+9x+3}{\sqrt{11y^2+(2y-3)(y-x)+3x}}+\frac{4x}{\sqrt{xy}+x} \right ]=0\Leftrightarrow y=x\)
Tới đây thay vào phương trình (2) giải dễ dàng.
giải hệ phương trình
\(\left\{{}\begin{matrix}\sqrt{x-2}+\sqrt{y-3}=3\\2\sqrt{x-2}-3\sqrt{y-3}=-4\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{3x}{x+1}+\dfrac{2}{y+4}=4\\\dfrac{2x}{x+1}-\dfrac{5}{y+4}=4\end{matrix}\right.\)
a.
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge2\\y\ge3\end{matrix}\right.\)
\(\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\2\sqrt{x-2}-3\sqrt{y-3}=-4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\5\sqrt{x-2}=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\\sqrt{x-2}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=1\\\sqrt{y-3}=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=7\end{matrix}\right.\)
b.
ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\y\ne-4\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{15x}{x+1}+\dfrac{10}{y+4}=20\\\dfrac{4x}{x+1}-\dfrac{10}{y+4}=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{15x}{x+1}+\dfrac{10}{y+4}=20\\\dfrac{19x}{x+1}=28\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{x+1}=\dfrac{28}{19}\\\dfrac{1}{y+4}=-\dfrac{4}{19}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}19x=28x+28\\4y+16=-19\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{28}{9}\\y=-\dfrac{35}{4}\end{matrix}\right.\)
Giải hệ phương trình:
1, \(\left\{{}\begin{matrix}x^2+1+y^2+xy=y\\x+y-2=\frac{y}{1+x^2}\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}x^3+8y^3-4xy^2=1\\2x^4+8y^4-2x-y=0\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}x^2+y^2=\frac{1}{5}\\4x^2+3x-\frac{57}{25}=-y\left(3x+1\right)\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}\sqrt{12-y}+\sqrt{y\left(12-x\right)}=12\\x^3-8x-1=2\sqrt{y-2}\end{matrix}\right.\)
5, \(\left\{{}\begin{matrix}\left(1-y\right)\sqrt{x-y}+x=2+\left(x-y-1\right)\sqrt{y}\\2y^2-3x+6y+1=2\sqrt{x-2y}-\sqrt{4x-5y-3}\end{matrix}\right.\)
Giải hệ phương trình:
1. \(\left\{{}\begin{matrix}\sqrt{x}+2\sqrt{-1}=5\\4\sqrt{x}-\sqrt{y-1}=2\end{matrix}\right.\)
2. \(\left\{{}\begin{matrix}\sqrt{3x-1}-\sqrt{2y+1}=1\\2\sqrt{3x-1}+3\sqrt{2y+1}=12\end{matrix}\right.\)
3. \(\left\{{}\begin{matrix}\sqrt{x-2}+\sqrt{y-3}=3\\2\sqrt{x-2}-3\sqrt{y-3}=-4\end{matrix}\right.\)
4. \(\left\{{}\begin{matrix}2\sqrt{x+1}-3\sqrt{-2}=5\\4\sqrt{x+1}+\sqrt{y-2}=17\end{matrix}\right.\)
2) Ta có: \(\left\{{}\begin{matrix}\sqrt{3x-1}-\sqrt{2y+1}=1\\2\sqrt{3x-1}+3\sqrt{2y+1}=12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{3x-1}-2\sqrt{2y+1}=2\\2\sqrt{3x-1}+3\sqrt{2y+1}=12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-5\sqrt{2y+1}=-10\\\sqrt{3x-1}-\sqrt{2y+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2y+1}=2\\\sqrt{3x-1}-2=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2y+1=4\\3x-1=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2y=3\\3x=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{3}{2}\\x=\dfrac{10}{3}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{10}{3}\\y=\dfrac{3}{2}\end{matrix}\right.\)
3) Ta có: \(\left\{{}\begin{matrix}\sqrt{x-2}+\sqrt{y-3}=3\\2\sqrt{x-2}-3\sqrt{y-3}=-4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x-2}+2\sqrt{y-3}=6\\2\sqrt{x-2}-3\sqrt{y-3}=-4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5\sqrt{y-3}=10\\\sqrt{x-2}+\sqrt{y-3}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{y-3}=2\\\sqrt{x-2}+2=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y-3=4\\x-2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=7\\x=3\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=3\\y=7\end{matrix}\right.\)
1) Giải hệ phương trình
\(\left\{{}\begin{matrix}3x^2+xy-4x+2y=2\\x\left(x+1\right)+y\left(y+1\right)=4\end{matrix}\right.\)
2) Giải phương trình
\(\sqrt{x^2-5x+4}+2\sqrt{x+5}=2\sqrt{x-4}+\sqrt{x^2+4x-5}\)
3) Tính giá trị của biểu thức
\(A=2x^3+3x^2-4x+2\)
Với \(x=\sqrt{2+\sqrt{\dfrac{5+\sqrt{5}}{2}}}+\sqrt{2-\sqrt{\dfrac{5+\sqrt{5}}{2}}}-\sqrt{3-\sqrt{5}}-1\)
4) Cho x, y thỏa mãn:
\(\sqrt{x+2014}+\sqrt{2015-x}-\sqrt{2014-x}=\sqrt{y+2014}+\sqrt{2015-y}-\sqrt{2014-y}\)
Chứng minh \(x=y\)
Câu 4:
Giả sử điều cần chứng minh là đúng
\(\Rightarrow x=y\), thay vào điều kiện ở đề bài, ta được:
\(\sqrt{x+2014}+\sqrt{2015-x}-\sqrt{2014-x}=\sqrt{x+2014}+\sqrt{2015-x}-\sqrt{2014-x}\) (luôn đúng)
Vậy điều cần chứng minh là đúng
2) \(\sqrt{x^2-5x+4}+2\sqrt{x+5}=2\sqrt{x-4}+\sqrt{x^2+4x-5}\)
⇔ \(\sqrt{\left(x-4\right)\left(x-1\right)}-2\sqrt{x-4}+2\sqrt{x+5}-\sqrt{\left(x+5\right)\left(x-1\right)}=0\)
⇔ \(\sqrt{x-4}.\left(\sqrt{x-1}-2\right)-\sqrt{x+5}\left(\sqrt{x-1}-2\right)=0\)
⇔ \(\left(\sqrt{x-4}-\sqrt{x+5}\right)\left(\sqrt{x-1}-2\right)=0\)
⇔ \(\left[{}\begin{matrix}\sqrt{x-4}-\sqrt{x+5}=0\\\sqrt{x-1}-2=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}\sqrt{x-4}=\sqrt{x+5}\\\sqrt{x-1}=2\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x\in\varnothing\\x=5\end{matrix}\right.\)
⇔ x = 5
Vậy S = {5}
Bài 1:
ĐKĐB suy ra $x(x+1)+y(y+1)=3x^2+xy-4x+2y+2$
$\Leftrightarrow 2x^2+x(y-5)+(y-y^2+2)=0$
Coi đây là PT bậc 2 ẩn $x$
$\Delta=(y-5)^2-4(y-y^2+2)=(3y-3)^2$Do đó:
$x=\frac{y+1}{2}$ hoặc $x=2-y$. Thay vào một trong 2 phương trình ban đầu ta thu được:
$(x,y)=(\frac{-4}{5}, \frac{-13}{5}); (1,1)$
giải hệ phương trình (theo 4 cách):
a/ \(\left\{{}\begin{matrix}\sqrt{5}x-y=\sqrt{5}\left(\sqrt{3}-1\right)\\2\sqrt{3}x+3\sqrt{5}y=21\end{matrix}\right.\)
b/ \(\left\{{}\begin{matrix}1,7x-2y=3,8\\2,1x+5y=0,4\end{matrix}\right.\)
a: \(\left\{{}\begin{matrix}\sqrt{5}x-y=\sqrt{5}\left(\sqrt{3}-1\right)\\2\sqrt{3}x+3\sqrt{5}y=21\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2\sqrt{15}x-2\sqrt{3}\cdot y=2\sqrt{15}\left(\sqrt{3}-1\right)\\2\sqrt{15}x+15y=21\sqrt{5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-2\sqrt{3}y-15y=2\sqrt{45}-2\sqrt{15}-21\sqrt{5}\\2\sqrt{3}x+3\sqrt{5}y=21\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y\left(-2\sqrt{3}-15\right)=-15\sqrt{5}-2\sqrt{15}\\2\sqrt{3}\cdot x+3\sqrt{5}\cdot y=21\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{15\sqrt{5}+2\sqrt{15}}{2\sqrt{3}+15}=\sqrt{5}\\2\sqrt{3}x+3\sqrt{5}\cdot y=21\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\sqrt{5}\\2\sqrt{3}x=21-3\sqrt{5}\cdot\sqrt{5}=21-15=6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\sqrt{5}\\x=\dfrac{6}{2\sqrt{3}}=\sqrt{3}\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}1,7x-2y=3,8\\2,1x+5y=0,4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}8,5x-10y=19\\4,2x+10y=0,8\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}8,5x-10y+4,2x+10y=19,8\\2,1x+5y=0,4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}12,7x=19,8\\2,1x+5y=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{198}{127}\\5y=0,4-2,1x=-\dfrac{365}{127}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{198}{127}\\y=-\dfrac{73}{127}\end{matrix}\right.\)
giải các hệ pt sau:
a) \(\left\{{}\begin{matrix}x+2y=-1\\x-y=5\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\frac{5}{x}-\frac{6}{y}=3\\\frac{4}{x}+\frac{9}{y}=7\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}3\sqrt{x+1}+\sqrt{y-1}=1\\\sqrt{x+1}-\sqrt{y-1}=-2\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}\left|x-1\right|+y=5\\4x+3y=23\end{matrix}\right.\)
a) \(\left\{{}\begin{matrix}x+2y=-1\\x-y=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3y=-6\\x-y=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-2\\x=3\end{matrix}\right.\)
Vậy..............................................................................
b) \(\left\{{}\begin{matrix}\frac{5}{x}-\frac{6}{y}=3\\\frac{4}{x}+\frac{9}{y}=7\end{matrix}\right.\)ĐKXĐ: x,y≠0
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{20}{x}-\frac{24}{y}=12\\\frac{20}{x}+\frac{45}{y}=35\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\frac{69}{y}=23\\\frac{20}{x}+\frac{45}{y}=35\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=3\\x=10\end{matrix}\right.\)
Vậy...................................................................................
c) \(\left\{{}\begin{matrix}3\sqrt{x+1}+\sqrt{y-1}=1\\\sqrt{x+1}-\sqrt{y-1}=-2\end{matrix}\right.\)ĐKXĐ:\(\left\{{}\begin{matrix}x\ge-1\\y\ge1\end{matrix}\right.\)
\(\Rightarrow4\sqrt{x+1}\)\(=-1\)(vô nghiệm)
Vậy hệ pt vô nghiệm
d) Nhân 3 pt đầu rồi thu gọn
Giải hệ phương trình
\(\left\{{}\begin{matrix}2x^3-4x^2+3x-1=2x^3\left(2-y\right)\sqrt{3-2y}\\\left(\sqrt{x\sqrt{3-2y}-\sqrt{x}}\right)^2\left(\sqrt{x\sqrt{3-2y}+2}+\sqrt{x+1}\right)=4\end{matrix}\right.\)
bài này mình chưa giải dc triệt để ở cái cuối
\(2x^3-4x^2+3x-1=2x^3\left(2-y\right)\sqrt{3-2y}\) \(\left(y\le\dfrac{3}{2}\right)\)
\(\Leftrightarrow4x^3-8x^2+6x-2=2x^3\left(4-2y\right)\sqrt{3-2y}\left(1\right)\)
\(đặt:\sqrt{3-2y}=a\ge0\Rightarrow a^2+1=4-2y\)
\(\left(1\right)\Leftrightarrow4x^3-8x^2+6x-2=2x^3.\left(a^2+1\right)a\)
\(\Leftrightarrow4x^3-8x^2+6x-2-2x^3\left(a^2+1\right)a\)
\(\Leftrightarrow-2\left(xa-x+1\right)\left[\left(xa\right)^2+x^2a+2x^2-xa-2x+1\right]=0\)
\(\Leftrightarrow x.a-x+1=0\Leftrightarrow x\left(a-1\right)=-1\Leftrightarrow x=\dfrac{-1}{a-1}\)
\(\left(\sqrt{x\sqrt{3-2y}-\sqrt{x}}\right) ^2=x\sqrt{3-2y}-\sqrt{x}\)
\(=\dfrac{-a}{a-1}-\sqrt{\dfrac{-1}{a-1}}\)
\(\left(\sqrt{x\sqrt{3-2y}+2}+\sqrt{x+1}\right)=\sqrt{\dfrac{-a}{a-1}+2}+\sqrt{\dfrac{a-2}{a-1}}\)
\(\Rightarrow\left(\dfrac{-a}{a-1}-\sqrt{-\dfrac{1}{a-1}}\right)\left(\sqrt{\dfrac{-a}{a-1}+2}+\sqrt{\dfrac{a-2}{a-1}}\right)-4=0\)
\(\Leftrightarrow\left(-\dfrac{a}{a-1}-\sqrt{-\dfrac{1}{a-1}}\right).2\sqrt{\dfrac{a-2}{a-1}}=4\)
\(\Leftrightarrow\left(-\dfrac{a}{a-1}-\sqrt{-\dfrac{1}{a-1}}\right)\sqrt{\dfrac{a-2}{a-1}}=2\)
\(\Leftrightarrow\left(-1+\dfrac{-1}{a-1}-\sqrt{-\dfrac{1}{a-1}}\right)\sqrt{1-\dfrac{1}{a-1}}=2\)(3)
\(đặt:1-\dfrac{1}{a-1}=u\Rightarrow\sqrt{-\dfrac{1}{a-1}}=\sqrt{u-1}\)
\(\left(3\right)\Leftrightarrow\left(u-2-\sqrt{u-1}\right)\sqrt{u}=2\)
bình phương lên tính được u
\(\Rightarrow u=.....\Rightarrow a\Rightarrow y=...\Rightarrow x=....\)
Với \(x=0\) không phải nghiệm
Với \(x>0\) chia 2 vế cho pt đầu cho \(x^3\)
\(\Rightarrow2-\dfrac{4}{x}+\dfrac{3}{x^2}-\dfrac{1}{x^3}=2\left(2-y\right)\sqrt{3-2y}\)
\(\Leftrightarrow1-\dfrac{1}{x}+\left(1-\dfrac{1}{x}\right)^3=\sqrt{3-2y}+\sqrt{\left(3-2y\right)^3}\)
Xét hàm \(f\left(t\right)=t+t^3\Rightarrow f'\left(t\right)=1+3t^2>0\Rightarrow f\left(t\right)\) đồng biến
\(\Rightarrow1-\dfrac{1}{x}=\sqrt{3-2y}\)
Thế vào pt dưới:
\(\left(\sqrt{x\left(1-\dfrac{1}{x}\right)-\sqrt{x}}\right)^2\left(\sqrt{x\left(1-\dfrac{1}{x}\right)+2}+\sqrt{x+1}\right)=4\)
\(\Leftrightarrow\left(x-\sqrt{x}-1\right)\left(\sqrt{x+1}+\sqrt{x+1}\right)=4\)
\(\Leftrightarrow\left(x-\sqrt{x}-1\right)\sqrt{x+1}=2\)
Phương trình này ko có nghiệm đẹp, chắc bạn ghi nhầm đề bài của pt dưới
... giải ra \(1-\dfrac{1}{x}=\sqrt{3-2y}\)
Thế xuống pt dưới:
\(\left(\sqrt{x\left(1-\dfrac{1}{x}\right)+2}\right)^2\left(\sqrt{x\left(1-\dfrac{1}{x}\right)+2}+\sqrt{x-1}\right)^4=4\)
\(\Leftrightarrow\left(x+1\right)\left(\sqrt{x+1}+\sqrt{x-1}\right)^4=4\)
Có vẻ đề bài vẫn sai
Do \(x\ge1\) theo ĐKXĐ nên \(x+1\ge2\) ; \(\left(\sqrt{x+1}+\sqrt{x-1}\right)^4\ge\left(\sqrt{2}+0\right)^4=4\)
\(\Rightarrow\left(x+1\right)\left(\sqrt{x+1}+\sqrt{x-1}\right)^4\ge8>4\) nên pt vô nghiệm