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\(f\left(x\right)+h\left(x\right)-g\left(x\right)\)
\(=\left(5x^4+3x^2+x-1\right)+\left(-x^4+3x^3-2x^2-x+2\right)\)
\(-\left(2x^4-x^3+x^2+2x+1\right)\)
\(=\left(5x^4-x^4-2x^4\right)+\left(3x^3+x^3\right)+\left(3x^2-2x^2-x^2\right)\)
\(+\left(x-x-2x\right)+\left(-1+2-1\right)\)
\(=2x^4+4x^3-2x\)
a, \(f\left(x\right)=2x^2\left(x-1\right)-5\left(x+2\right)-2x\left(x-2\right)\)
\(=2x^3-2x^2-5x-10-2x^2+4x=2x^3-4x^2-x-10\)
b, \(g\left(x\right)=x^2\left(2x-3\right)-x\left(x+1\right)-\left(3x-2\right)\)
\(=2x^3-3x^2-x^2-x-3x+2=2x^3+2-4x^2-4x\)
b, Ta có : \(H\left(x\right)=F\left(x\right)-G\left(x\right)=2x^3-4x^2-x-10-2x^3+4x^2+4x-2\)
\(\Leftrightarrow3x-12=0\Leftrightarrow x=4\)
Ta có: \(F\left(x\right)+G\left(x\right)-H\left(x\right)=0\)
\(\Leftrightarrow4x^2+3x-2+3x^2-2x+5-5x^2+2x-3=0\\ \Leftrightarrow2x^2+3x=0\\ \Rightarrow x\left(2x+3\right)=0\\ \Rightarrow x=0;x=\dfrac{-3}{2}\)
Vậy tìm được x thỏa mãn là: \(x=0;x=\dfrac{-3}{2}\)
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
Bài này chill ha ? nhưng ko ai lm cx lạ :vvv
a, Ta có : \(f\left(1\right)=5.1-1^3+3.1^2-1=5-1+3-1=6\)
\(g\left(-1\right)=-\left(-1\right)^3+3\left(-1\right)^2+2\left(-1\right)-3=1+3-2-3=-1\)
\(f\left(1\right)-g\left(-1\right)=6-\left(-1\right)=7\)
b, Ta có :
\(h\left(x\right)=f\left(x\right)-g\left(x\right)=\left(5x-x^3+3x^2-1\right)-\left(-x^3+3x^2+2x-3\right)\)
\(=5x-x^3+3x^2-1+x^3-3x^2-2x+3=3x+2\)
c, \(\left|h\left(x\right)-5\right|+2x=2,5\Leftrightarrow\left|3x+2-5\right|+2x=2,5\)
\(\Leftrightarrow\left|3x-3\right|+2x=2,5\Leftrightarrow\left|3x-3\right|=2,5-2x\)
Chia 2 TH nhá vì lười :3 (nhưng ko dám chắc nha men)
Ta có \(f\left(1\right)=g\left(2\right)\)
hay \(2.1^2+a.1+4=2^2-5.2-b\)
\(2+a+4\) \(=4-10-b\)
\(6+a\) \(=-6-b\)
\(a+b\) \(=-6-6\)
\(a+b\) \(=-12\) \(\left(1\right)\)
Lại có \(f\left(-1\right)=g\left(5\right)\)
hay \(2.\left(-1\right)^2+a.\left(-1\right)+4=5^2-5.5-b\)
\(2-a+4\) \(=25-25-b\)
\(6-a\) \(=-b\)
\(-a+b\) \(=-6\)
\(b-a\) \(=-6\)
\(b\) \(=-b+a\) \(\left(2\right)\)
Thay \(\left(2\right)\) vào \(\left(1\right)\) ta được:
\(a+\left(-6+a\right)=-12\)
\(a-6+a\) \(=-12\)
\(a+a\) \(=-12+6\)
\(2a\) \(=-6\)
\(a\) \(=-6:2\)
\(a\) \(=-3\)
Mà \(a=-3\)
⇒ \(b=-6+\left(-3\right)=-9\)
Vậy \(a=3\) và \(b=-9\)
Cái Vậy \(a=3\) và \(b=-9\) bạn ghi là \(a=-3\) và \(b=-9\) nha mk quên ghi dấu " \(-\) "
a) f (x) + h (x) = g (x)
⇒h(x)=g(x)−f(x)⇒h(x)=g(x)−f(x)
h(x)=(x4−x3+x2+5)−(x4−3x2+x−1)h(x)=(x4−x3+x2+5)−(x4−3x2+x−1)
h(x)=x4−x3+x2+5−x4+3x2−x+1h(x)=−x3+4x2−x+6h(x)=x4−x3+x2+5−x4+3x2−x+1h(x)=−x3+4x2−x+6
b) f (x) - h (x) = g (x)
⇒h(x)=f(x)−g(x)⇔h(x)=(x4−3x2+x−1)−(x4−x3+x2+5)⇒h(x)=f(x)−g(x)⇔h(x)=(x4−3x2+x−1)−(x4−x3+x2+5)
⇔h(x)=x4−3x2+x−1−x4+x3−x2−5⇔h(x)=x3−4x2+x−6
a. Ta có: f(x) + h(x) = g(x)
Suy ra: h(x) = g(x) – f(x) = (x4 – x3 + x2 + 5) – (x4 – 3x2 + x – 1)
= x4 – x3 + x2 + 5 – x4 + 3x2 – x + 1
= -x3 + 4x2 – x + 6
b. Ta có: f(x) – h(x) = g(x)
Suy ra: h(x) = f(x) – g(x) = (x4 – 3x2 + x – 1) – (x4 – x3 + x2 + 5)
= x4 – 3x2 + x – 1 – x4 + x3 – x2 – 5
= x3 – 4x2 + x – 6
a) \(f\left(x\right)-g\left(x\right)+h\left(x\right)\)
\(=x^3-2x^2+3x+1-\left(x^3+x-1\right)+\left(2x^2-1\right)\)
\(=x^3-2x^2+3x+1-x^3-x+1+2x^2-1\)
\(=2x+1\)
b) \(f\left(x\right)-g\left(x\right)+h\left(x\right)=0\)
\(\Leftrightarrow\)\(2x+1=0\)
\(\Leftrightarrow\)\(x=-\frac{1}{2}\)