A(x) ở đâu

Đề có gì đó hơi sai sai 

a) Ta có: A(x)=M(x)-N(x)

\(=\dfrac{5}{2}x^2-\dfrac{1}{2}x-x^3-1-\left(-2x^3+5x^2-12+4x\right)\)

\(=\dfrac{5}{2}x^2-\dfrac{1}{2}x-x^3-1+2x^3-5x^2+12-4x\)

\(=x^3-\dfrac{5}{2}x^2-\dfrac{9}{2}x+11\)

Cái chỗ 1;1/2 là gì vậy bạn?

a) N(x)= -2x³ + 5x² -12 +2x

M(x)= -x³ + 2,5x² - 0.5x -1

-

N(x)= -2x³ + 5x² + 2x - 12

=

A(x)=M(x) - N(x)= x³ - 2,5x² -2,5x +11

b) M(x) = -x³ + 2,5x² - 0,5x -1

+

N(x) = -2x³ + 5x² + 2x -12

=

B(x)= M(x) + N(x) = -3x³ + 7,5x² + 1,5x -13

⇒ Bậc của B(x) là 6

`M+N= 0,5x^4 -4x^3 +2x-2,5 + 2x^3 +x^2+1,5`

`= 0,5x^4 +(-4x^3+ 2x^3 ) +x^2+2x +(-2,5 +1,5)`

`= 0,5x^4 -2x^3 +x^2+2x -1`

\(M+N=0,5x^4-4x^3+2x-2,5+2x^3+x^2+1,5\)

\(=0,5x^4-4x^3+2x^3+x^2+2x-2,5+1,5\)

\(=0,5x^4-2x^3+x^2+2x-1\)

\(M+N=\left(0,5x^4-4x^3+2x-2,5\right)+\left(2x^3+x^2+1,5\right)\)
\(=0,5x^4-4x^3+2x-2,5+2x^3+x^2+1,5\)
\(=0,5x^4-\left(4x^3-2x^3\right)+x^2+2x-\left(2,5-1,5\right)\)
\(=0,5x^4-2x^3+x^2+2x-1\)

1)\(\left(4x-10\right)\left(24+5x\right)=0\)

\(\Leftrightarrow2\left(2x-5\right)\left(24+5x\right)=0\)

Vì 2≠0

nên \(\left[{}\begin{matrix}2x-5=0\\24+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\5x=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{-24}{5}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{5}{2};\frac{-24}{5}\right\}\)

2) \(0,5x\left(x-3\right)=\left(x-3\right)\left(2,5x-4\right)\)

\(\Leftrightarrow0,5x\left(x-3\right)-\left(x-3\right)\left(2,5x-4\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left[0,5x-\left(2,5x-4\right)\right]=0\)

\(\Leftrightarrow\left(x-3\right)\left(0,5x-2,5x+4\right)=0\)

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

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

\(\Leftrightarrow\left(x-3\right)\cdot2\cdot\left(2-x\right)=0\)

Vì 2≠0

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

Vậy: x∈{2;3}

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

\(\Leftrightarrow\left(2x+1\right)\left(2x-1\right)-\left(2x+1\right)\left(3x-5\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left[2x-1-\left(3x-5\right)\right]=0\)

\(\Leftrightarrow\left(2x+1\right)\left(2x-1-3x+5\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left(4-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\4-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-1\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=4\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{-1}{2};4\right\}\)

4) \(\left(2-3x\right)\left(x+11\right)=\left(3x-2\right)\left(2-5x\right)\)

\(\Leftrightarrow\left(2-3x\right)\left(x+11\right)-\left(3x-2\right)\left(2-5x\right)=0\)

\(\Leftrightarrow\left(2-3x\right)\left(x+11\right)+\left(2-3x\right)\left(2-5x\right)=0\)

\(\Leftrightarrow\left(2-3x\right)\left(x+11+2-5x\right)=0\)

\(\Leftrightarrow\left(2-3x\right)\left(13-4x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2-3x=0\\13-4x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\4x=13\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=\frac{13}{4}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{2}{3};\frac{13}{4}\right\}\)

`@` `\text {Answer}`

`A = -3x^2 - 0,5x^2 + 2,5x^2`

`= (-3 - 0,5 + 2,5)x^2`

`= (-3,5+2,5)x^2`

`= -x^2`

Hệ số: `-1`

Phần biến: `x`

Bậc: `2`

`b)`

Thay `x = -1/2`

`- (-1/2)^2`

`= - (1/4)`

`= -1/4`

Vậy, `A = -1/4.`

`7,`

`a,`

\(M(x) = - 5x ^ 4 + 3x ^ 5 + x(x ^ 2 + 5) + 14x ^ 4 - 6x ^ 5 - x ^ 3 + x - 1 \)

\(M(x)=-5x^4+3x^5+x^3+5x+14x^4-6x^5-x^3+x-1\)

`M(x)=(3x^5-6x^5)+(-5x^4+14x^4)+(x^3-x^3)+(5x+x)-1`

`M(x)=-3x^5+9x^4+6x-1`

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

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

`N(x)=(x^5+2x^5)+(-5x^4-4x^4)+(-3x^3+3x^3)+3x-5`

`N(x)=3x^5-9x^4+3x-5`

`b,`

`H(x)=M(x)+N(x)`

\(H(x)=(-3x^5+9x^4+6x-1)+(3x^5-9x^4+3x-5) \)

`H(x)=-3x^5+9x^4+6x-1+3x^5-9x^4+3x-5`

`H(x)=(-3x^5+3x^5)+(9x^4-9x^4)+(6x+3x)+(-1-5)`

`H(x)=9x-6`

`G(x)=M(x)-N(x)`

\(G(x)=(-3x^5+9x^4+6x-1)-(3x^5-9x^4+3x-5)\)

`G(x)=-3x^5+9x^4+6x-1-3x^5+9x^4-3x+5`

`G(x)=(-3x^5-3x^5)+(9x^4+9x^4)+(6x-3x)+(-1+5)`

`G(x)=-6x^5+18x^4+3x+4`

`c,`

`H(x)=9x-6`

Hệ số cao nhất của đa thức: `9`

Hệ số tự do: `-6`

`G(x)=-6x^5+18x^4+3x+4`

Hệ số cao nhất của đa thức: `-6`

Hệ số tự do: `4`

`d,`

`H(-1)=9*(-1)-6=-9-6=-15`

`H(1)=9*1-6=9-6=3`

`G(1)=-6*1^5+18*1^4+3*1+4`

`G(1)=-6+18+3+4=12+3+4=15+4=19`

`G(0)=-6*0^5+18*0^4+3*0+4=4`

`H(-3/2)=9*(-3/2)-6=-27/2-6=-39/2`

`e,`

Đặt `H(x)=9x-6=0`

`-> 9x=0+6`

`-> 9x=6`

`-> x=6 \div 9`

`-> x=2/3`

Vậy, nghiệm của đa thức là `x=2/3.`