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

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

\(=x^3+3x^2-5x-15+\left(x+4\right)\left(x-x^2\right)\)

\(=x^3+3x^2-5x-15-x^3+x^2-4x^2+4x\)

\(=3x^2-5x-15-3x^2+4x\)

\(=-x-15\)

2) Đặt đa thức là \(N\left(x\right)\)ta được: \(3x^3+2x^2-x+k=N\left(x\right)\left(x-1\right)\)

Để \(3x^3+2x^2-x+K⋮x-1\Leftrightarrow x=1\)

Thay vào ta được

\(\Rightarrow3.1^3+2.1^2-1+K=0\)

\(\Rightarrow3+2-1+K=0\)

\(\Rightarrow K=-4\)

3) \(Q=x^2-x+3\)

\(=x^2-x+\frac{1}{4}+\frac{11}{4}\)

\(=\left(x-\frac{1}{2}\right)^2+\frac{11}{4}\)

Với \(\forall x\) ta có: \(\left(x-\frac{1}{2}\right)^2\ge0\Rightarrow N=\left(x-\frac{1}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}>0\)

nhanh giùm mình được không

Bài 1: 

a) Ta có: \(P=1+\dfrac{3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)\)

\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}-\dfrac{3x}{3\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)

\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{4}{x-2}-\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)

\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\dfrac{4\left(x+2\right)-x-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{4x+8-x-x+2}\)

\(=1+3\cdot\dfrac{\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)

\(=1+\dfrac{3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)

\(=\dfrac{\left(x+3\right)\left(2x+10\right)+3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)

\(=\dfrac{2x^2+10x+6x+30+3x-6}{\left(x+3\right)\left(2x+10\right)}\)

\(=\dfrac{2x^2+19x-6}{\left(x+3\right)\left(2x+10\right)}\)

Alo đề nghị viết đề một cách chính xác 

`@` `\text {Ans}`

`\downarrow`

Gửi c!

Bài 1: 

a) \(3x^2\left(2x^3-x+5\right)-6x^5-3x^3+10x^2\)

\(=6x^5-3x^3+10x^2-6x^5-3x^3+10x^2\)

\(=10x^2+10x^2\)

\(=20x^2\)

b) \(-2x\left(x^3-3x^2-x+11\right)-2x^4+3x^3+2x^2-22x\)

\(=-2x^4+6x^3+2x^2-22x-2x^4+3x^3+2x^2-22x\)

\(=-4x^4+9x^3+4x^2-44x\)

4:

a: =>1/4x^2-1/4x^2+2x=-14

=>2x=-14

=>x=-7

b: =>2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20

=>3x^2-12x-2=3x^2-17x+20

=>5x=22

=>x=22/5

a: \(A=x^3-27-x^3+3x^2-3x+1-4\left(x^2-4\right)-x\)

\(=3x^2-4x-26-4x^2+16\)

\(=-x^2-4x-10\)

a: \(x^2-5x+10\)

\(=x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}+\dfrac{15}{4}\)

\(=\left(x-\dfrac{5}{2}\right)^2+\dfrac{15}{4}>0\forall x\)

b: \(2x^2+8x+15\)

\(=2\left(x^2+4x+\dfrac{15}{2}\right)\)

\(=2\left(x^2+4x+4+\dfrac{7}{2}\right)\)

\(=2\left(x+2\right)^2+7>0\forall x\)

Ta có : 

\(x^2-4x+5=\left(x^2-2.2x+2^2\right)+1=\left(x-2\right)^2+1\ge1>0\)

Vậy đa thức \(x^2-4x+5\) vô nghiệm với mọi giá trị của x 

Chúc bạn học tốt ~