a,\(x\ge\dfrac{3}{2}\)

\(\dfrac{\sqrt{2x-3}}{\sqrt{x-1}}=2\)\(=>2\sqrt{x-1}=\sqrt{2x-3}\)

\(< =>4\left(x-1\right)=2x-3< =>4x-4=2x-3< =>x=0,5\left(ktm\right)\)

\(=>x\in\phi\)

b, \(đk:\left[{}\begin{matrix}x< 1\\x\ge\dfrac{3}{2}\end{matrix}\right.\)

\(=>\sqrt{\dfrac{2x-3}{x-1}}=4< =>\dfrac{2x-3}{x-1}=>4\left(x-1\right)=2x-3\)

\(< =>4x-4=2x-3< =>2x=1=>x=\dfrac{1}{2}\left(tm\right)\)

vậy,,,..

\(a,\) Đặt \(A=\dfrac{3x^2-2x+3}{x^2+1}\Leftrightarrow Ax^2+A=3x^2-2x+3\)

\(\Leftrightarrow x^2\left(A-3\right)-2x+A-3=0\)

Coi đây là PT bậc 2 ẩn x, PT có nghiệm 

\(\Leftrightarrow\Delta=4-4\left(A-3\right)^2\ge0\\ \Leftrightarrow\left(A-3\right)^2\le1\Leftrightarrow2\le A\le4\)

Vậy \(A_{min}=4\Leftrightarrow\dfrac{3x^2-2x+3}{x^2+1}=4\Leftrightarrow x=...\)

\(b,\) Đặt \(B=\dfrac{3x^2-4x+4}{x^2+2}\Leftrightarrow Bx^2+2B=3x^2-4x+4\)

\(\Leftrightarrow x^2\left(B-3\right)+4x+2B-4=0\)

Coi đây là PT bậc 2 ẩn x, PT có nghiệm

\(\Leftrightarrow\Delta=16-8\left(B-2\right)\left(B-3\right)\ge0\\ \Leftrightarrow\left(B-2\right)\left(B-3\right)\le2\\ \Leftrightarrow B^2-5B+4\le0\\ \Leftrightarrow\left(B-1\right)\left(B-4\right)\le0\\ \Leftrightarrow1\le B\le4\)

Vậy\(B_{min}=4\Leftrightarrow\dfrac{3x^2-4x+4}{x^2+2}=4\Leftrightarrow x=...\)

\(a,\Leftrightarrow2x^3-x^2+ax+b=\left(x-1\right)\left(x+1\right)\cdot a\left(x\right)\)

Thay \(x=1\Leftrightarrow2-1+a+b=0\Leftrightarrow a+b=-1\)

Thay \(x=-1\Leftrightarrow-2-1-a+b=0\Leftrightarrow b-a=3\)

Từ đó ta được \(\left\{{}\begin{matrix}a+b=-1\\-a+b=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-2\\b=1\end{matrix}\right.\)

\(b,\Leftrightarrow ax^3+bx^2+2x-1=\left(x-1\right)\left(x+6\right)\cdot b\left(x\right)\)

Thay \(x=1\Leftrightarrow a+b+2-1=0\Leftrightarrow a+b=-1\)

Thay \(x=-6\Leftrightarrow-216a+36b+12-1=0\Leftrightarrow216a-36b=11\)

Từ đó ta được \(\left\{{}\begin{matrix}a+b=-1\\216a-36b=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{25}{252}\\b=-\dfrac{227}{252}\end{matrix}\right.\)

\(c,\Leftrightarrow ax^4+bx^3+1=\left(x+1\right)^2\cdot c\left(x\right)\)

Thay \(x=-1\Leftrightarrow a-b+1=0\Leftrightarrow b=a+1\)

\(\Leftrightarrow ax^4+\left(a+1\right)x^3+1⋮\left(x+1\right)\\ \Leftrightarrow ax^4+ax^3+x^3+1⋮\left(x+1\right)\\ \Leftrightarrow ax^3\left(x+1\right)+\left(x+1\right)\left(x^2-x+1\right)⋮\left(x+1\right)\\ \Leftrightarrow\left(x+1\right)\left(ax^3+x^2-x+1\right)⋮\left(x+1\right)\\ \Leftrightarrow ax^3+x^2-x+1⋮\left(x+1\right)\)

Thay \(x=-1\Leftrightarrow-a+1+1+1=0\Leftrightarrow a=3\Leftrightarrow b=4\)

\(ĐK:x\ne\pm2;x\ne-1\\ PT\Leftrightarrow\dfrac{x^2-x-2+x^2+x-2}{\left(x+2\right)\left(x-2\right)}=\dfrac{2x+1}{x+1}\\ \Leftrightarrow\dfrac{2x^2-4}{x^2-4}=\dfrac{2x+1}{x+1}\\ \Leftrightarrow\left(2x^2-4\right)\left(x+1\right)=\left(x^2-4\right)\left(2x+1\right)\\ \Leftrightarrow2x^3+2x^2-4x-4=2x^3+x^2-8x-4\\ \Leftrightarrow x^2+4x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=-4\left(tm\right)\end{matrix}\right.\)

d: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{4}}=\dfrac{x+3y-2z}{\dfrac{1}{2}+3\cdot\dfrac{1}{3}-2\cdot\dfrac{1}{4}}=\dfrac{36}{1}=36\)

Do đó: x=18; y=12; z=9

a) Thay x + 3y - 2z vào biểu thức ta có:

 \(\dfrac{x - 1}{3} = \dfrac{3(y + 2)}{3 . 4} = \dfrac{2(z - 2)}{2 . 3}\) = \(​​​​\dfrac{x - 1}{3} = \dfrac{3x + 6}{12} = \dfrac{2z - 4}{6}\)

Áp dụng tính chất dãy tỉ số bằng nhua ta có:

\(​​​​\dfrac{x - 1}{3} = \dfrac{3y + 6}{12} = \dfrac{2z - 4}{6} = ​​​​\dfrac{x - 1}{3}+ \dfrac{3y + 6}{12} -\dfrac{2z - 4}{6}\) 

=\(​​​​\dfrac{x - 1 + 3y + 6 - 2z + 4}{3 + 12 -6} \) = \(​​​​\dfrac{(x + 3y - 2z) + ( -1 + 6 +4)}{3 + 12 - 6} \)

=\(​​​​\dfrac{36 + 9}{9}\) = 5

=> \(​​​​\dfrac{x - 1}{3} =\) 5 => x - 1 = 5.3 =15 => x = 5+1 = 6

=>

=>

Vậy ...

(Bạn dựa theo cách này và lm những bài tiếp nhé!)

b) \(2x^2+4y^2+z^2-4xy-2x-2z+5=0\)

\(\Leftrightarrow\left(x^2-4xy+4y^2\right)+\left(x^2-2x+1\right)+\left(z^2-2z+1\right)+3=0\)

....

a) \(x^2+5y^2-4xy+6y+9=0\)

\(\Leftrightarrow\left(x^2-4xy+4y^2\right)+\left(y^2+6y+9\right)=0\)

\(\Leftrightarrow\left(x-2y\right)^2+\left(y+3\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2y=0\\y+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2y=2.\left(-3\right)=-6\\y=-3\end{matrix}\right.\)

Vậy : \(\left(x,y\right)=\left(-6,-3\right)\)