Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)
f(x)=9-x5+4x-2x3+x2-7x4
= -x5-7x4-2x3+x2+4x+9
g(x)=x5-9+2x2+7x4+2x3-3x
=x5+7x4+2x3+2x2-3x-9
b)h(x)=(-x5-7x4-2x3+x2+4x+9)+(x5+7x4+2x3+2x2-3x-9)
= -x5+(-7x4)+(-2x3)+x2+4x+9+x5+7x4+2x3-3x+(-9)+2x2
=(-x5+x5)+(-7x4+7x4)+(-2x3+2x3)+(x2+2x2)+(4x-3x)+(-9+9)
= 3x2+x
c)h(x)=3x2+x
Ta có:3x2+x=0
x(3x+1)=0
TH1:x=0
TH2:3x+1=0
=>x=-1/3
Vậy=0 và -1/3 là nghiệm của h(x)
a. f(x)+g(x)=2x5−4x4+3x3−x2+5x−1+(−x5+2x4−3x3−x2−2x+7)
=2x5-x5-4x4+2x4+3x3-3x3-x2-x2+5x-2x-1+7
=x5-2x4-2x2+3x+6
b. f(x)+h(x)=2x5−4x4+3x3−x2+5x−1+x5−2x4−2x2−x−3
=2x5+x5-4x4-2x4+3x3-x2-2x2+5x-x-1-3
=3x5-6x4+3x3-3x2+6x-4
c. g(x)+h(x)=−x5+2x4−3x3−x2−2x+7+x5−2x4−2x2−x−3
=-x5+x5+2x4-2x4-3x3-x2-2x2-2x-x+7-3
=-3x3-3x2-3x+4
d. f(x)-g(x)=2x5−4x4+3x3−x2+5x−1-(−x5+2x4−3x3−x2−2x+7)
=2x5−4x4+3x3−x2+5x−1-x5-2x4+3x3+x2+2x-7
=2x5-x5-4x4-2x4+3x3+3x3-x2+x2+5x+2x-1-7
=x5-6x4+6x3+7x-8
e. f(x)-h(x)=2x5−4x4+3x3−x2+5x−1-(x5−2x4−2x2−x−3)
=2x5−4x4+3x3−x2+5x−1-x5+2x4+2x2+x+3
=2x5-x5-4x4+2x4+3x3-x2+2x2+5x+x-1+3
=x5-2x4+3x3+x2+6x-4
h. g(x)-h(x)=−x5+2x4−3x3−x2−2x+7-(x5−2x4−2x2−x−3)
=−x5+2x4−3x3−x2−2x+7-x5+2x4+2x2+x+3
=-x5-x5+2x4+2x4-3x3-x2+2x2-2x+x+7+3
=-2x5+4x4-3x3+x2-x+10
f. f(x)+g(x)+h(x)=2x5−4x4+3x3−x2+5x−1+(−x5+2x4−3x3−x2−2x+7)+x5−2x4−2x2−x−3
=2x5-x5+x5-4x4+2x4-2x4+3x3-3x3-x2-x2-2x2+5x-2x-x-1+7-3
=2x5-4x4-4x2+2x+3
g. f(x)+g(x)-h(x)=2x5−4x4+3x3−x2+5x−1+(−x5+2x4−3x3−x2−2x+7)-(x5−2x4−2x2−x−3)
=2x5−4x4+3x3−x2+5x−1+(−x5+2x4−3x3−x2−2x+7)-x5+2x4+2x2+x+3
=2x5-x5-x5-4x4+2x4+2x4+3x3-3x3-x2-x2+2x2+5x-2x+x-1+7+3
=4x+9
n. f(x)-g(x)+h(x)=2x5−4x4+3x3−x2+5x−1-(−x5+2x4−3x3−x2−2x+7)+x5−2x4−2x2−x−3
=2x5−4x4+3x3−x2+5x−1-x5-2x4+3x3+x2+2x-7+x5−2x4−2x2−x−3
=2x5-x5+x5-4x4-2x4-2x4+3x3+3x3-x2+x2-2x2+5x+2x-x-1-7-3
=2x5-8x4+6x3-2x2+6x-11
m. f(x)-g(x)-h(x)=2x5−4x4+3x3−x2+5x−1-(−x5+2x4−3x3−x2−2x+7)-(x5−2x4−2x2−x−3)
=2x5−4x4+3x3−x2+5x−1-x5-2x4+3x3+x2+2x-7-x5+2x4+2x2+x+3
=2x5-x5-x5-4x4-2x4+2x4+3x3+3x3-x2+x2+2x2+5x+2x+x-1-7+3
=-4x4+6x3+2x2+8x-5
\(a,x.\frac{-3}{7}=\frac{4}{21}\)
\(x=\frac{4}{21}:\frac{-3}{7}\)
\(x=\frac{-4}{9}\)
\(b,\frac{-4}{7}:x=\frac{2}{5}\)
\(x=\frac{-4}{7}:\frac{2}{5}\)
\(x=\frac{-10}{7}\)
\(c,x+\frac{1}{12}=\frac{-3}{8}\)
\(x=\frac{-3}{8}-\frac{1}{12}\)
\(x=\frac{-11}{24}\)
\(d,\frac{2}{15}-x=\frac{-3}{10}\)
\(x=\frac{2}{15}+\frac{3}{10}\)
\(x=\frac{13}{30}\)
\(e,-x+\frac{4}{5}=\frac{1}{2}\)
\(-x=\frac{-3}{10}\)
\(x=\frac{3}{10}\)
\(f,\frac{3}{4}.\left(x+1\right)-\frac{1}{2}=\frac{3}{7}\)
\(\frac{3}{4}.\left(x+1\right)=\frac{13}{14}\)
\(x+1=\frac{26}{21}\)
\(x=\frac{5}{21}\)
\(\frac{-3}{2}-2x+\frac{3}{4}=-2\)
\(\frac{-3}{2}-2x=\frac{-11}{4}\)
\(2x=\frac{-3}{2}+\frac{11}{4}\)
\(2x=\frac{-17}{4}\)
\(x=\frac{-17}{8}\)
\(h,-x+\frac{4}{5}=\frac{1}{2}\)
\(-x=\frac{-3}{10}\)
\(x=\frac{3}{10}\)
chúc bạn học tốt !!!
Bài 1:
a; \(\dfrac{7}{8}\) + \(x\) = \(\dfrac{4}{7}\)
\(x\) = \(\dfrac{4}{7}\) - \(\dfrac{7}{8}\)
\(x\) = \(\dfrac{32}{56}\) - \(\dfrac{49}{56}\)
\(x=-\) \(\dfrac{49}{56}\)
Vậy \(x=-\dfrac{49}{56}\)
b; 6 - \(x\) = - \(\dfrac{3}{4}\)
\(x\) = 6 + \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{24}{4}+\dfrac{3}{4}\)
\(x=\dfrac{27}{4}\)
Vậy \(x=\dfrac{27}{4}\)
c; \(\dfrac{1}{-5}\) + \(x\) = \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{3}{4}\) + \(\dfrac{1}{5}\)
\(x=\dfrac{15}{20}\) + \(\dfrac{4}{20}\)
\(x=\dfrac{19}{20}\)
Vậy \(x=\dfrac{19}{20}\)
Bài 1:
d; - 6 - \(x\) = - \(\dfrac{3}{5}\)
\(x\) = - 6 + \(\dfrac{3}{5}\)
\(x=-\dfrac{30}{5}\) + \(\dfrac{3}{5}\)
\(x=-\dfrac{27}{5}\)
Vậy \(x=-\dfrac{27}{5}\)
e; - \(\dfrac{2}{6}\) + \(x\) = \(\dfrac{5}{7}\)
\(x\) = \(\dfrac{5}{7}\) + \(\dfrac{2}{6}\)
\(x\) = \(\dfrac{15}{21}\) + \(\dfrac{1}{3}\)
\(x=\dfrac{15}{21}\) + \(\dfrac{7}{21}\)
\(x=\dfrac{22}{21}\)
Vậy \(x=\dfrac{22}{21}\)
f; - 8 - \(x\) = - \(\dfrac{5}{3}\)
\(x\) = \(-\dfrac{5}{3}\) + 8
\(x\) = \(\dfrac{-5}{3}\) + \(\dfrac{24}{3}\)
\(x\) = \(\dfrac{-19}{3}\)
Vậy \(x=-\dfrac{19}{3}\)
a: \(h\left(x\right)=7x^5+x^4-2x^3+4+x^4+6x^3-9x^2-2x-1=7x^5+2x^4+4x^3-9x^2-2x+3\)
b: \(h\left(x\right)=7x^5+x^4-2x^3+4-x^4-6x^3+9x^2+2x+1=7x^5-8x^3+9x^2+2x+5\)
\(P\left(x\right)+Q\left(x\right)=f\left(x\right)-g\left(x\right)\)
\(f\left(x\right)-g\left(x\right)=3x^4+3x^3-5x^2+x-5-x^4-3x^3+3x^2-5x+7\)
\(=2x^4-2x^2-4x+2\)
\(\Rightarrow P\left(x\right)+Q\left(x\right)=2x^4-2x^2-4x+2\left(1\right)\)
\(P\left(x\right)-Q\left(x\right)=g\left(x\right)+h\left(x\right)\)
\(g\left(x\right)+h\left(x\right)=x^4+3x^3-3x^2+5x-7+5x^4+2x^3+x^2-5\)
\(=6x^4+5x^3-2x^2+5x-12\)
\(\Rightarrow P\left(x\right)-Q\left(x\right)=6x^4+5x^3-2x^2+5x-12\left(2\right)\)
Từ ( 1 );( 2 ) thì tìm dc P(x) và Q(x)
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
