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a) \(f\left(x\right)=5x^3-7x^2+x+7+4x^5\)
\(f\left(-1\right)=5.\left(-1\right)^3-7.\left(-1\right)^2+\left(-1\right)+7+4.\left(-1\right)^5\)
\(f\left(-1\right)=\left(-5\right)-7+\left(-1\right)+7+\left(-4\right)\)
\(f\left(-1\right)=-10\)
\(\Rightarrow f\left(x\right)=-10\)
\(g\left(x\right)=4x^5-3x^3-7x^2+2x+5\)
\(g\left(0\right)=4.0^5-3.0^3-7.0^2+2.0+5\)
\(g\left(0\right)=5\)
\(\Rightarrow g\left(x\right)=0\)
\(h\left(x\right)=x^2-4x-5\)
\(h\left(-\frac{1}{2}\right)=\left(-\frac{1}{2}\right)^2-4.\left(-\frac{1}{2}\right)-5\)
\(h\left(-\frac{1}{2}\right)=\frac{1}{4}-\left(-2\right)-5\)
\(h\left(-\frac{1}{2}\right)=-\frac{11}{4}\)
\(\Rightarrow h\left(x\right)=-\frac{11}{4}\)
\(f\left(-1\right)=5\left(-1\right)^3-7\left(-1\right)^2+\left(-1\right)+7+4\left(-1\right)^5\)
\(f\left(-1\right)=-5-7-1+7-4\)
\(f\left(-1\right)=-10\)
\(g\left(0\right)=4.0^5-3.0^3-7.0^2+2.0+5\)
\(g\left(0\right)=0-0-0+0+5\)
\(g\left(0\right)=5\)
\(h\left(-\frac{1}{2}\right)=\left(-\frac{1}{2}\right)^2-4\left(-\frac{1}{2}\right)-5\)
\(h\left(-\frac{1}{2}\right)=\frac{1}{4}-\left(-2\right)-5\)
\(h\left(-\frac{1}{2}\right)=\frac{1}{4}+2-5\)
\(h\left(-\frac{1}{2}\right)=-\frac{11}{4}\)
a: \(f\left(x\right)+g\left(x\right)-h\left(x\right)\)
\(=5x^5-4x^4+3x^3-x^2-3x+4+x^5-2x^4+x^3-x+7\)
\(=6x^5-6x^4+4x^3-x^2-4x+11\)
f(x)-g(x)-h(x)
\(=15x^5-12x^4+9x^3-7x^2+7x+x^5-2x^4+x^3-x+7\)
\(=16x^5-14x^4+10x^3-7x^2+6x+7\)
b: f(x)+2g(x)=0
\(\Leftrightarrow10x^5-8x^4+6x^3-4x^2+2x+2-10x^5+8x^4-6x^3+6x^2-10x+4=0\)
\(\Leftrightarrow2x^2-8x+6=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
=>x=1 hoặc x=3
bài 1
a) \(-\frac{1}{3}xy\).(3\(x^2yz^2\))
=\(\left(-\frac{1}{3}.3\right)\).\(\left(x.x^2\right)\).(y.y).\(z^2\)
=\(-x^3\).\(y^2z^2\)
b)-54\(y^2\).b.x
=(-54.b).\(y^2x\)
=-54b\(y^2x\)
c) -2.\(x^2y.\left(\frac{1}{2}\right)^2.x.\left(y^2.x\right)^3\)
=\(-2x^2y.\frac{1}{4}.x.y^6.x^3\)
=\(\left(-2.\frac{1}{4}\right).\left(x^2.x.x^3\right).\left(y.y^2\right)\)
=\(\frac{-1}{2}x^6y^3\)
Bài 3:
a) \(f\left(x\right)=-15x^2+5x^4-4x^2+8x^2-9x^3-x^4+15-7x^3\)
\(f\left(x\right)=\left(5x^4-x^4\right)-\left(9x^3+7x^3\right)-\left(15x^2+4x^2-8x^2\right)+15\)
\(f\left(x\right)=4x^4-16x^3-11x^2+15\)
b)
\(f\left(x\right)=4x^4-16x^3-11x^2+15\)
\(f\left(1\right)=4\cdot1^4-16\cdot1^3-11\cdot1^2+15\)
\(f\left(1\right)=4\cdot1^4-16\cdot1^3-11\cdot1^2+15\)
\(f\left(1\right)=-8\)
\(f\left(x\right)=4x^4-16x^3-11x^2+15\)
\(f\left(-1\right)=4\cdot\left(-1\right)^4-16\cdot\left(-1\right)^3-11\cdot\left(-1\right)^2+15\)
\(f\left(-1\right)=24\)
a) Ta có: \(x^3-x^2+x-1=0\)
\(\Rightarrow x^2\left(x-1\right)+\left(x-1\right)=0\)
\(\Rightarrow\left(x^2+1\right)\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2+1=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x^2=-1\left(loại\right)\\x=1\end{matrix}\right.\)
Vậy x = 1 là nghiệm của đa thức f(x)
b, c: @Ace Legona
a)\(f\left(x\right)=x^3-x^2+x-1\)
Cho \(f\left(x\right)=0\Rightarrow x^3-x^2+x-1=0\)
\(\Rightarrow x^2\left(x-1\right)+\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x^2+1\right)=0\)
Dễ thấy: \(x^2+1\ge1>0\forall x\) ( vô nghiệm )
\(\Rightarrow x-1=0\Rightarrow x=1\)
b)\(g\left(x\right)=11x^3+5x^2+4x+10\)
Cho \(g\left(x\right)=0\Rightarrow11x^3+5x^2+4x+10=0\)
\(\Rightarrow11x^3-6x^2+10x+11x^2-6x+10=0\)
\(\Rightarrow x\left(11x^2-6x+10\right)+\left(11x^2-6x+10\right)=0\)
\(\Rightarrow\left(x+1\right)\left(11x^2-6x+10\right)=0\)
Dễ thấy:
\(11x^2-6x+10=11\left(x-\dfrac{3}{11}\right)^2+\dfrac{101}{11}\ge\dfrac{101}{11}>0\forall x\) (vô nghiệm)
\(\Rightarrow x+1=0\Rightarrow x=-1\)
c)\(h\left(x\right)=-17x^3+8x^2-3x+12\)
Cho \(h\left(x\right)=0\Rightarrow-17x^3+8x^2-3x+12=0\)
\(\Rightarrow17x^2+9x+12-17x^3-9x^2-12x=0\)
\(\Rightarrow\left(17x^2+9x+12\right)-x\left(17x^2+9x+12\right)=0\)
\(\Rightarrow\left(1-x\right)\left(17x^2+9x+12\right)=0\)
Dễ thấy:
\(17x^2+9x+12=17\left(x+\dfrac{9}{34}\right)^2+\dfrac{735}{68}\ge\dfrac{735}{68}>0\forall x\)(vô nghiệm)
\(\Rightarrow1-x=0\Rightarrow x=1\)
f(x) + g(x) - h(x) = (x5 - 4x3 + x2 - 2x + 1) + (x5 - 2x4 + x2 - 5x + 3) - (x4 - 3x2 + 2x - 5)
= x5 - 4x3 + x2 - 2x + 1 + x5 - 2x4 + x2 - 5x + 3 - x4 + 3x2 - 2x + 5
= (x5 + x5) - (2x4 + x4) - 4x3 + ( x2 + x2 + 3x2) - (2x + 5x + 2x) + (1 + 3 + 5)
= 2x5 - 3x4 - 4x3 + 5x2 - 9x + 9
f(x)=
f(x) + g(x) - h(x) = (x5 - 4x3 + x2 - 2x + 1) + (x5 - 2x4 + x2 - 5x + 3) - (x4 - 3x2 + 2x - 5)
= x5 - 4x3 + x2 - 2x + 1 + x5 - 2x4 + x2 - 5x + 3 - x4 + 3x2 - 2x + 5
= (x5 + x5) - (2x4 + x4) - 4x3 + ( x2 + x2 + 3x2) - (2x + 5x + 2x) + (1 + 3 + 5)
= 2x5 - 3x4 - 4x3 + 5x2 - 9x + 9
a, Ta có : \(f\left(x\right)-g\left(x\right)=h\left(x\right)\)hay
\(4x^2+3x+1-3x^2+2x-1=h\left(x\right)\)
\(\Rightarrow h\left(x\right)=x^2+5x\)
b, Đặt \(h\left(x\right)=x^2+5x=0\Leftrightarrow x\left(x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)
Vậy nghiệm của đa thức h(x) là x = -5 ; x = 0
Đặt \(k\left(x\right)=7x^2-35x+42=0\)
\(\Leftrightarrow7\left(x^2+5x+6\right)=0\)
\(\Leftrightarrow7\left(x^2+2x+3x+6\right)=0\Leftrightarrow7\left(x+2\right)\left(x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-3\end{cases}}\)
Vậy nghiệm của đa thức k(x) là x = -3 ; x = -2
xin lỗi mọi người 1 tý nha cái phần c) ý ạ đề thì vậy như thế nhưng có cái ở phần biểu thức ở dưới ý là
\(\left(\frac{3^2}{6}-81\right)^3\) chuyển thành \(\left(\frac{3^3}{6}81\right)^3\)
bị sai mỗi thế thôi ạ mọi người giúp em với ạ