a/ \(\left\{{}\begin{matrix}2x>-1\\4x\ge3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x>-\frac{1}{2}\\x\ge\frac{3}{4}\end{matrix}\right.\) \(\Rightarrow x\ge\frac{3}{4}\)

b/ \(x-1-\frac{1}{x-1}\ge0\)

\(\Leftrightarrow\frac{\left(x-1\right)^2-1}{x-1}\ge0\)

\(\Leftrightarrow\frac{x\left(x-2\right)}{x-1}\ge0\Rightarrow\left[{}\begin{matrix}x\ge2\\0\le x< 1\end{matrix}\right.\)

a) x⁴+x³+2x²+x+1=(x⁴+x³+x²)+(x²+x+1)=x²(x²+x+1)+(x²+x+1)=(x²+x+1)(x²+1)

b)a³+b³+c³-3abc=a³+3ab(a+b)+b³+c³ -(3ab(a+b)+3abc)=(a+b)³+c³-3ab(a+b+c)

=(a+b+c)((a+b)²-(a+b)c+c²)-3ab(a+b+c)=(a+b+c)(a²+2ab+b²-ac-ab+c²-3ab)=(a+b+c)(a²+b²+c²-ab-ac-bc)

c)Đặt x-y=a;y-z=b;z-x=c

a+b+c=x-y-z+z-x=o

đưa về như bài b

d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung

e)x²(y-z)+y²(z-x)+z²(x-y)=x²(y-z)-y²((y-z)+(x-y))+z²(x-y)

=x²(y-z)-y²(y-z)-y²(x-y)+z²(x-y)=(y-z)(x²-y²)-(x-y)(y²-z²)=(y-z)(x²-2y²+xy+xz+yz)

Ta có : \(\frac{3x-1}{x+2}\ge1\)

=> \(3x-1\ge x-2\)

=> \(3x-x\ge-2+1\)

=> \(x\ge-\frac{1}{2}\)

\(\frac{3\text{x}-1}{x-1}-\frac{2\text{x}+5}{x+3}=1-\)\(\frac{4}{x^2+2\text{x}-3}\)                              \(\left(\text{Đ}K\text{X}\text{Đ}:x\ne1;x\ne-3\right)\)

\(\Leftrightarrow\frac{\left(3\text{x}-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{\left(2\text{x}+5\right)\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}=\frac{\left(x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{4}{\left(x-1\right)\left(x+3\right)}\)

\(\Rightarrow\left(3\text{x}-1\right)\left(x+3\right)-\left(2\text{x}+5\right)\left(x-1\right)=\left(x-1\right)\left(x+3\right)-4\)

\(\Leftrightarrow3\text{x}^2+8\text{x}-3-2\text{x}^2-3\text{x}+5=x^2+2\text{x}-3-4\)

\(\Leftrightarrow3\text{x}^2-2\text{x}^2-x^2+8\text{x}-3\text{x}-2\text{x}=-3-4+3-5\Leftrightarrow3\text{x}=-9\Leftrightarrow x=-3\)(không thỏa mãn ĐKXĐ)

Vậy pt vô nghiệm

ĐKXĐ: \(x\ne1;x\ne2\)

\(\frac{2x-1}{x-1}+\frac{3x-2}{x-2}=\frac{x^2+4x+5}{x^2-3x+2}+4\\ \Leftrightarrow\frac{2x^2-5x+2}{\left(x-1\right)\left(x-2\right)}+\frac{3x^2-5x+2}{\left(x-1\right)\left(x-2\right)}=\frac{x^2+4x+5}{\left(x-1\right)\left(x-2\right)}+\frac{4x^2-12x+8}{\left(x-1\right)\left(x-2\right)}\\ \Rightarrow2x^2-5x+2+3x^2-5x+2=x^2+4x+5+4x^2-12x+8\\ \Leftrightarrow2x^2+3x^2-x^2-4x^2-5x-5x-4x+12x=5-2-2\\ \Leftrightarrow-2x=1\\ \Leftrightarrow x=\frac{-1}{2}\left(tm\right)\)Vậy tập nghiệm của phương trình là: \(S=\left\{-\frac{1}{2}\right\}\)

ĐKXĐ: x∉{1;2}

Ta có: \(\frac{2x-1}{x-1}+\frac{3x-2}{x-2}=\frac{x^2+4x+5}{x^2-3x+2}+4\)

\(\Leftrightarrow\frac{\left(2x-1\right)\left(x-2\right)}{\left(x-1\right)\left(x-2\right)}+\frac{\left(3x-2\right)\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}=\frac{x^2+4x+5}{\left(x-1\right)\left(x-2\right)}+\frac{4\left(x^2-3x+2\right)}{\left(x-1\right)\left(x-2\right)}\)

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

\(\Leftrightarrow2x^2-4x-x+2+3x^2-3x-2x+2=x^2+4x+5+4x^2-12x+8\)

\(\Leftrightarrow5x^2-10x+4=5x^2-8x+13\)

\(\Leftrightarrow5x^2-10x+4-5x^2+8x-13=0\)

\(\Leftrightarrow-2x-9=0\)

\(\Leftrightarrow-2x=9\)

hay \(x=\frac{-9}{2}\)(tm)

Vậy: \(S=\left\{-\frac{9}{2}\right\}\)

\(a,4x-6< 7x-12\)

\(\Leftrightarrow6< 3x\Leftrightarrow x>2\)

\(b,\frac{3x-7}{4}\ge2-\frac{x+5}{3}\)

\(\Leftrightarrow3\left(3x-7\right)\ge24-4\left(x+5\right)\)

\(\Leftrightarrow13x\ge25\Leftrightarrow x\ge\frac{25}{13}\)

\(c,\frac{3x-8}{-7}\ge1-\frac{x+2}{-3}\)

\(\Leftrightarrow-3\left(3x-8\right)\ge21+7\left(x+2\right)\)

\(\Leftrightarrow-16x\ge11\)

\(\Leftrightarrow x\le-\frac{11}{16}\)

\(d,-12-8x>3+2x-\left(5-7x\right)\)

\(\Leftrightarrow14>17x\Leftrightarrow x< \frac{14}{17}\)

\(e,-1+\frac{x-1}{-3}\le\frac{x+2}{-9}\)

\(\Leftrightarrow-9-3\left(x-1\right)\le-\left(x+2\right)\)

\(\Leftrightarrow-2x\le4\Leftrightarrow x\ge-2\)

a.\(\Leftrightarrow\left(x+3\right)\left(x^2-x-2-2x^2+3x+5\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=3\\x=-1\end{matrix}\right.\)

(x-2)(x+1)(x+3)=(x+3)(x+1)(2x-58)

\(x^3+2x^2-5x-6\)=\(2x^3+3x^2-14x-15\)

\(-x^3-x^2+9x+9=0\)

\(-x^2\left(x+1\right)+9\left(x+1\right)=0\)

\(\left(x+1\right)\left(9-x^2\right)\)=0

(x+1)(3-x)(3+x)=0

*x+1=0 =>x=-1

*3-x=0=>x=3

*3+x=0=>x=-3

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

ĐK: \(x\ge0\)

\(\sqrt{2x+1}-\sqrt{3x}=3x-\left(2x+1\right)\)

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

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

\(\Leftrightarrow\sqrt{2x+1}=\sqrt{3x}\Rightarrow x=1\left(tm\right)\)

ai giải hộ mk ý a vs ý c

c) \(x^3y+xy^3-3x^2-3y^2=17\)

\(\Leftrightarrow xy\left(x^2+y^2\right)-3\left(x^2+y^2\right)=17\Leftrightarrow\left(x^2+y^2\right)\left(xy-3\right)=17\)

\(\Leftrightarrow\left(x^2+y^2\right),\left(xy-3\right)\inƯ\left(17\right)\)

Do \(x^2+y^2\ge0\Rightarrow x^2+y^2\in\left\{1;17\right\}\)

TH1:  \(\hept{\begin{cases}x^2+y^2=1\\xy-3=17\end{cases}}\Rightarrow\hept{\begin{cases}\frac{400}{y^2}+y^2=1\\x=\frac{20}{y}\end{cases}}\) (vô nghiệm)

TH2:  \(\hept{\begin{cases}x^2+y^2=17\\xy-3=1\end{cases}}\Rightarrow\hept{\begin{cases}\frac{16}{y^2}+y^2=17\\x=\frac{4}{y}\end{cases}}\)

Ta có bảng:

Vậy các cặp số nguyên thỏa mãn là (x;y) = (1;4) ; (-1;-4) ; (4;1) ; (-4;-1).