ĐKXĐ: \(x\ge\dfrac{2}{3}\)

\(\Leftrightarrow x\sqrt{3x-2}-x^2+\left(x+1\right)\sqrt{5x-1}-\left(x+1\right)^2+x^2+\left(x+1\right)^2-8x+3=0\)

\(\Leftrightarrow x\left(\sqrt{3x-2}-x\right)+\left(x+1\right)\left(\sqrt{5x-1}-x-1\right)+2\left(x^2-3x+2\right)=0\)

\(\Leftrightarrow\dfrac{-x\left(x^2-3x+2\right)}{\sqrt{3x-2}+x}+\dfrac{-\left(x+1\right)\left(x^2-3x+2\right)}{\sqrt{5x-1}+x+1}+2\left(x^2-3x+2\right)=0\)

\(\Leftrightarrow\left(x^2-3x+2\right)+\left(2-\dfrac{x}{\sqrt{3x-2}+x}-\dfrac{x+1}{\sqrt{5x-1}+x+1}\right)=0\)

\(\Leftrightarrow\left(x^2-3x+2\right)\left(\dfrac{\sqrt{3x-2}}{\sqrt{3x-2}+x}+\dfrac{\sqrt{5x-1}}{\sqrt{5x-1}+x+1}\right)=0\)

\(\Leftrightarrow x^2-3x+2=0\) (ngoặc đằng sau luôn dương)

\(\Leftrightarrow...\)

Câu 1:

\(2x^3-3x^2+x+a\)

\(=2\left(x^3-6x^2+12x-8\right)+9\left(x^2-4x+4\right)+13\left(x-2\right)+\left(6+a\right)\)

\(=2\left(x-2\right)^3+9\left(x-2\right)^2+13\left(x-2\right)+\left(6+a\right)\)chia hết cho \(x-2\)khi và chỉ khi :

\(6+a=0\Leftrightarrow a=-6\). Vậy \(a=-6\).

Câu 2:

\(\left(x+1\right)\left(2x-x\right)-\left(3x+5\right)\left(x+2\right)=4x^2+1\)

\(\Leftrightarrow x^2+x-\left(3x^2+11x+10\right)=-4x^2+1\)

\(\Leftrightarrow x^2+x-3x^2-11x-10+4x^2-1=0\)

\(\Leftrightarrow2x^2-10x-11=0\)

\(\Delta'=\left(-5\right)^2-2\left(-11\right)=47>0\)

\(\Rightarrow\)Phương trình có 2 nghiệm phân biệt:

\(x=\frac{5+\sqrt{47}}{2}\)hoặc \(x=\frac{5-\sqrt{47}}{2}\)

Vậy phương trình có tập nghiệm \(S=\left\{\frac{5+\sqrt{47}}{2};\frac{5-\sqrt{47}}{2}\right\}\)

Ta có : \(\frac{2x+5}{x+1}=\frac{2x+2+3}{x+1}=\frac{2\left(x+1\right)+3}{x+1}=2+\frac{3}{x+1}\)

Vì 2 \(\inℤ\Rightarrow\frac{3}{x+1}\inℤ\Rightarrow3⋮x+1\Rightarrow x+1\inƯ\left(3\right)\Rightarrow x+1\in\left\{1;3;-1;-3\right\}\)

=> \(x\in\left\{0;2-2;-4\right\}\)

Để \(\frac{3x-1}{2x-1}\inℤ\Rightarrow3x-1⋮2x-1\Rightarrow2\left(3x-1\right)⋮2x-1\Rightarrow6x-2⋮2x-1\)

=> \(6x-3+1⋮2x-1\Rightarrow3\left(2x-1\right)+1⋮2x-1\)

Vì \(3\left(2x-1\right)⋮2x-1\)

=> \(1⋮2x-1\Rightarrow2x-1\inƯ\left(1\right)\Rightarrow2x-1\in\left\{1;-1\right\}\Rightarrow x\in\left\{1;0\right\}\)

\(\frac{2x+5}{x+1}=\frac{2\left(x+1\right)+3}{x+1}=2+\frac{3}{x+1}\)

Để phân số nguyên => \(\frac{3}{x+1}\)nguyên

=> \(3⋮x+1\)

=> \(x+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

=> \(x=\left\{0;-2;2;-4\right\}\)

\(\frac{3x-1}{2x-1}\)

Để phân số nguyên => \(3x-1⋮2x-1\)

=> \(2\left(3x-1\right)⋮2x-1\)

=> \(6x-2⋮2x-1\)

\(\Rightarrow3\left(2x-1\right)+1⋮2x-1\)

\(\Rightarrow1⋮2x-1\)

\(\Rightarrow2x-1\inƯ\left(1\right)=\left\{\pm1\right\}\)

\(\Rightarrow x=\left\{1;0\right\}\)

a, \(\frac{2x+5}{x+1}=\frac{2\left(x+1\right)+3}{x+1}=\frac{3}{x+1}\)

\(\Leftrightarrow x+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

1.\(3x^2+12x-66=0\)

\(\Rightarrow\)\(3\left(x^2+4x+4\right)-78=0\)

\(\Rightarrow3\left(x+2\right)^2=78\)

\(\Rightarrow\left(x+2\right)^2=26\)

\(\Rightarrow x+2=\sqrt{26}\)hoặc \(x+2=-\sqrt{26}\)

\(\Rightarrow x=\sqrt{26}-2\)hoặc \(x=-\sqrt{26}-2\)

7: Ta có: \(\left(3x+4\right)\left(2x-1\right)+6x\left(1-x\right)=0\)

\(\Leftrightarrow6x^2-3x+8x-4+6x-6x^2=0\)

\(\Leftrightarrow11x=4\)

hay \(x=\dfrac{4}{11}\)

8: Ta có: \(2x\left(x^2-1\right)+x\left(-2x^2-3x+1\right)=-x-27\)

\(\Leftrightarrow2x^3-2x-2x^3-3x^2+x+x+27=0\)

\(\Leftrightarrow x^2=9\)

hay \(x\in\left\{3;-3\right\}\)

Bạn là thảo linh hoc lớp 6a1 dung k

\(\frac{1}{4}+\frac{1}{3}:3x=-5\)

\(\frac{1}{3}:3x=(-5)-\frac{1}{4}\)

\(\frac{1}{3}:3x=\frac{-21}{4}\)

\(\frac{1}{9}\cdot x=\frac{-21}{4}\)

\(x=\frac{-21}{4}:\frac{1}{9}\)

x=\(\frac{-189}{4}\)

Vậy x=\(\frac{-189}{4}\)

\(\frac{1}{4}+\frac{1}{3}:3x=-5\Rightarrow\frac{1}{3}:3x=\left(-5\right)-\frac{1}{4}=\frac{-21}{4}\)

\(3x=\frac{1}{3}:\frac{-21}{4}=\frac{1}{3}.\frac{4}{21}=\frac{4}{63}\)

\(\Rightarrow x=\frac{4}{63}:3=\frac{4}{63}.\frac{1}{3}=\frac{4}{189}\)

Vì \(\hept{\begin{cases}3x=5y\\2y=-3z\end{cases}}\Rightarrow\hept{\begin{cases}\frac{x}{5}=\frac{y}{3}\\\frac{y}{-3}=\frac{z}{2}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}\frac{x}{5}=\frac{y}{3}\\\frac{y}{3}=\frac{z}{-2}\end{cases}}\Rightarrow\frac{x}{5}=\frac{y}{3}=\frac{z}{-2}\)

Áp dụng tính chất dãy tỉ số bằng nhau , ta có :

\(\frac{x}{5}=\frac{y}{3}=\frac{z}{-2}=\frac{x+y-z}{5+3-\left(-2\right)}=\frac{2}{10}=\frac{1}{5}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{1}{5}.5=1\\y=\frac{1}{5}.3=\frac{3}{5}\\z=\frac{1}{5}.\left(-2\right)=\frac{-2}{5}\end{cases}}\)

Ta có : 

\(3x=5y\Leftrightarrow\frac{x}{5}=\frac{y}{3}\)

\(2y=-3z\Leftrightarrow\frac{y}{3}=-\frac{z}{2}\)

Do đó : 

\(\frac{x}{5}=\frac{y}{3}=-\frac{z}{2}\)

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

\(\frac{x}{5}=\frac{y}{3}=-\frac{z}{2}=\frac{x+y-z}{5+3-2}=\frac{2}{6}=\frac{1}{3}\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{5}=\frac{1}{3}\Rightarrow x=\frac{5}{3}\\\frac{y}{3}=\frac{1}{3}\Rightarrow y=1\\-\frac{z}{2}=\frac{1}{3}\Rightarrow-\frac{2}{3}\end{cases}}\)

Vậy ... 

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

\(\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{3x-2y}{3\cdot5-2\cdot2}=\dfrac{44}{11}=4\)

Do đó: x=20; y=8

bn dẻo miệng quá!! 

Theo đề, ta có HPT:

\(\left\{{}\begin{matrix}2x=5y\\3x-2y=44\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-5y=0\\3x-2y=44\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x-15y=0\\6x-4y=88\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}11y=88\\2x=5y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=8\\x=20\end{matrix}\right.\)