Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có: P(x) = -5x3 - 1313 + 8x4 + x2 và Q(x) = x2 – 5x – 2x3 + x4 - 2323.
Ta sắp xếp hai đa thức theo lũy thừa giảm dần của biến như sau:
.
Ta có: P(x) = -5x3 – 1/3 + 8x4 + x2 và Q(x) = x2 – 5x – 2x3 + x4 – 2/3.
Ta sắp xếp hai đa thức theo lũy thừa giảm dần của biến như sau:
`@` `\text {Ans}`
`\downarrow`
`P(x)+Q(x)-R(x)`
`= 5x^2 + 5x - 4 +2x^2 - 3x + 1 - (4x^2 - x + 3)`
`= 5x^2 + 5x - 4 + 2x^2 - 3x + 1 - 4x^2 + x - 3`
`= (5x^2 + 2x^2 - 4x^2) + (5x - 3x + x) + (-4 + 1 - 3)`
`= 3x^2 + 3x - 6`
Thay `x=-1/2`
`3*(-1/2)^2 + 3*(-1/2) - 6`
`= 3*1/4 - 3/2 - 6`
`= 3/4 - 3/2 - 6`
`= -3/4 - 6 = -27/4`
Vậy, khi `x=-1/2` thì GTr của đa thức là `-27/4`
P(x)+Q(x)-R(x)
=5x^2+5x-4+2x^2-3x+1-4x^2+x-3
=2x^2+3x-6(1)
Khi x=-1/2 thì (1) sẽ là 2*1/4+3*(-1/2)-6=1/2-3/2-6=-7
a/ \(\left(x+3\right)\left(3\left(x^2+1\right)^2+2\left(x+3\right)^2\right)=5\left(x^2+1\right)^3\)
\(\Leftrightarrow3\left(x+3\right)\left(x^2+1\right)^2+2\left(x+3\right)^3-5\left(x^2+1\right)^3=0\)
\(\Leftrightarrow3\left(x+3\right)\left(x^2+1\right)^2-3\left(x^2+1\right)^3+2\left(x+3\right)^3-2\left(x^2+1\right)^3=0\)
\(\Leftrightarrow3\left(x^2+1\right)^2\left(-x^2+x+2\right)+2\left(-x^2+x+2\right)\left(\left(x+3\right)^2+\left(x+3\right)\left(x^2+1\right)+\left(x^2+1\right)^2\right)=0\)
\(\Leftrightarrow\left(-x^2+x+2\right)\left[3\left(x^2+1\right)^2+2\left(x+3+\dfrac{x^2+1}{2}\right)^2+\dfrac{3\left(x^2+1\right)^2}{4}\right]=0\)
\(\Leftrightarrow-x^2+x+2=0\) (phần ngoặc phía sau luôn dương)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
b/ \(3\left(x^2+2x-1\right)^2-2\left(x^2+3x-1\right)^2+5\left(x^2+3x-1-\left(x^2+2x-1\right)\right)^2=0\)
Đặt \(\left\{{}\begin{matrix}a=x^2+2x-1\\b=x^2+3x-1\end{matrix}\right.\)
\(3a^2-2b^2+5\left(b-a\right)^2=0\Leftrightarrow8a^2+3b^2-10ab=0\)
\(\Leftrightarrow\left(4a-3b\right)\left(2a-b\right)=0\Leftrightarrow\left[{}\begin{matrix}4a=3b\\2a=b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4\left(x^2+2x-1\right)=3\left(x^2+3x-1\right)\\2\left(x^2+2x-1\right)=x^2+3x-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-x-1=0\\x^2+x-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{2}\\x=\dfrac{1-\sqrt{5}}{2}\\x=\dfrac{-1+\sqrt{5}}{2}\\x=\dfrac{-1-\sqrt{5}}{2}\end{matrix}\right.\)
a: \(\Leftrightarrow x^3-3x^2+3x-1-x^3+2x^2-x=5x\left(2-x\right)-11\left(x+2\right)\)
=>-x^2+2x-1=10x-5x^2-11x-22
=>-x^2+2x-1=-5x^2-x-22
=>4x^2+3x+21=0
=>PTVN
b: \(\Leftrightarrow\left(x+10\right)\left(x+4\right)+3\left(x+4\right)\left(x-2\right)=4\left(x+10\right)\left(x-2\right)\)
=>x^2+14x+40+3(x^2+2x-8)=4(x^2+8x-20)
=>x^2+14x+40+3x^2+6x-24=4x^2+32x-80
=>20x+16=32x-80
=>-12x=-96
=>x=8
c: \(\Leftrightarrow6\left(x-3\right)+7\left(x-5\right)=13x+4\)
=>6x-18+7x-35=13x+4
=>-53=4(loại)
d: =>3(2x-1)-5(x-2)=3(x+7)
=>6x-3-5x+10=3x+21
=>3x+21=x+7
=>x=-7
e: =>x^3-6x^2+12x-8-x^3-3x^2-3x-1=-9x^2+1
=>-9x^2+9x-9=-9x^2+1
=>9x=10
=>x=10/9
1) \(A\left(x\right)=-5x^3+3x^4+\frac{5}{7}-8x^2-10x\)
\(A\left(x\right)=3x^4-5x^3-8x^2-10x+\frac{5}{7}\)
\(B\left(x\right)=-2x^4-\frac{2}{7}+7x^2+8x^3+6x\)
\(B\left(x\right)=-2x^4+8x^3+7x^2+6x-\frac{2}{7}\)
2) \(A\left(x\right)=3x^4-5x^3-8x^2-10x+\frac{5}{7}\)
+
\(B\left(x\right)=-2x^4+8x^3+7x^2+6x-\frac{2}{7}\)
\(A\left(x\right)+B\left(x\right)=x^4+3x^3-x^2-4x+\frac{3}{7}\)
\(A\left(x\right)=3x^4-5x^3-8x^2-10x+\frac{5}{7}\)
-
\(B\left(x\right)=-2x^4+8x^3+7x^2+6x-\frac{2}{7}\)
\(A\left(x\right)-B\left(x\right)=5x^4-13x^3-15x^2-16x+1\)
P(x)=-5x^3-1/3+8x^4+x^2
Q(x)=x^4-2x^3+x^2-5x-2/3
P(x)+Q(x)
=x^4-2x^3+x^2-5x-2/3+8x^4-5x^3+x^2-1/3
=9x^4-7x^3+2x^2-5x-1
P(x)-Q(x)
=x^4-2x^3+x^2-5x-2/3-8x^4+5x^3-x^2+1/3
=-7x^4+3x^3-5x-1/3