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\(A\left(x\right)+B\left(x\right)=3x^4-\frac{3}{4}x^3+2x^2-3+8x^4+\frac{1}{5}x^3-9x+\frac{2}{5}\)
\(=11x^4-\frac{11}{20}x^3+2x^2-\frac{13}{5}-9x\)
\(A\left(x\right)-B\left(x\right)=3x^4-\frac{3}{4}x^3+2x^2-3-8x^4-\frac{1}{5}x^3+9x-\frac{2}{5}\)
\(=-5x^4-\frac{19}{20}x^3+2x^2-\frac{17}{5}+9x\)
Bn làm nót nhé, tương tự thôi
\(A\left(x\right)+B\left(x\right)\)
\(=\left(3x^4-\frac{3}{4}x^3+2x^2-3\right)+\left(8x^4+\frac{1}{5}x^3-9x+\frac{2}{5}\right)\)
\(=11x^4-\frac{11}{20}x^3+2x^2-9x-\frac{13}{5}\)
\(A\left(x\right)-B\left(x\right)\)
\(=3x^4-\frac{3}{4}x^3+2x^2-3-8x^4-\frac{1}{5}x^3+9x-\frac{2}{5}\)
\(=-5x^4-\frac{19}{20}x^3+2x^2+9x-\frac{17}{5}\)
\(B\left(x\right)-A\left(x\right)\)
\(=8x^4+\frac{1}{5}x^3-9x+\frac{2}{5}-3x^4+\frac{3}{4}x^3+2x^2-3\)
\(=5x^4+\frac{19}{20}x^3+2x^2-9x-\frac{13}{5}\)
Bài 5:
a: \(P\left(x\right)=3x^5+x^4-2x^2+2x\)
\(Q\left(x\right)=-3x^5+2x^2-2x+3\)
b: \(P\left(x\right)+Q\left(x\right)=3x^5-3x^5+x^4-2x^2+2x^2+2x-2x+3\)
\(=x^4+3\)
\(P\left(x\right)-Q\left(x\right)=3x^5+x^4-2x^2+2x+3x^5-2x^2+2x-3\)
\(=6x^5+x^4-4x^2+4x-3\)
c: \(P\left(0\right)=3\cdot0^5+0^4-2\cdot0^2+2\cdot0=2\)
\(Q\left(0\right)=-3\cdot0^5+2\cdot0^2-2\cdot0+3=3\)
Vậy: x=0 là nghiệm của P(x), không là nghiệm của Q(x)
f(x) + g(x)
= (x5 - 3x2 + 7x4 - 9x3 + x2 - 1/4x) + (5x4 - x5 +x2 - 2x3 + 3x2 - 1/4)
= x5 - 3x2 + 7x4 - 9x3 + x2 - 1/4x + 5x4 - x5 +x2 - 2x3 + 3x2 - 1/4
=12x4 - 11x3 + 2x2 - 1/4x - 1/4
f(x) - g(x)
= (x5 - 3x2 + 7x4 - 9x3 + x2 - 1/4x) - (5x4 - x5 +x2 - 2x3 + 3x2 - 1/4)
= = x5 - 3x2 + 7x4 - 9x3 + x2 - 1/4x - 5x4 + x5 - x2 + 2x3 - 3x2 + 1/4
= 2x5 + 2x4 - 7x3 - 6x2 - 1/4x + 1/4
a)Sắp xếp : \(f\left(x\right)=x^5+7x^4-9x^3-2x^2-\dfrac{1}{4}x\)
\(g\left(x\right)=-x^5+5x^4-2x^3+4x^2-\dfrac{1}{4}x\)
Ta có : \(f\left(x\right)+g\left(x\right)=x^5+7x^4-9x^3-2x^2-\dfrac{1}{4}x-x^5+5x^4-2x^3+4x^2-\dfrac{1}{4}x\)
\(=12x^4-11x^3+2x^2-\dfrac{1}{2}x\)
\(f\left(x\right)-g\left(x\right)=x^5+7x^4-9x^3-2x^2-\dfrac{1}{4}x+x^5-5x^4+2x^3-4x^2+\dfrac{1}{4}x\)
\(=2x^5+2x^4-7x^3-6x^2\)
`@` `\text {Đáp án}`
`\downarrow`
`a,`
`A(x)+B(x)=`\(\left(3x^4-\dfrac{3}{4}x^3+2x^2-3\right)+8x^4+\dfrac{1}{5}x^3-9x+\dfrac{2}{5}\)
`= 3x^4-3/4x^3+2x^2-3+8x^4+1/5x^3-9x+2/5`
`= (3x^4+8x^4)+(-3/4x^3+1/5x^3)+2x^2-9x+(-3+2/5)`
`= 11x^4-11/20x^3+2x^2-9x-13/5`
`b,`
`A(x)-B(x)=`\(3x^4-\dfrac{3}{4}x^3+2x^2-3-\left(8x^4+\dfrac{1}{5}x^3-9x+\dfrac{2}{5}\right)\)
`=3x^4-3/4x^3+2x^2-3-8x^4-1/5x^3+9x-2/5`
`= (3x^4-8x^4)+(-3/4x^3-1/5x^3)+2x^2+9x+(-3-2/5)`
`= -5x^4 -19/20x^3+2x^2+9x-17/5`
`c,`
`B(x)-A(x)=`\(8x^4+\dfrac{1}{5}x^3-9x+\dfrac{2}{5}-\left(3x^4-\dfrac{3}{4}x^3+2x^2-3\right)\)
`= 8x^4+1/5x^3-9x+2/5 - 3x^4+3/4x^3-2x^2+3`
`= (8x^4-3x^4)+(1/5x^3-3/4x^3)-2x^2-9x+(2/5+3)`
`= 5x^4 + 19/20x^3 -2x^2 -9x+17/5`
a: A(x)+B(x)=11x^4-11/20x^3+2x^2-9x-13/5
b: A(x)-B(x)=-5x^4-19/20x^3+2x^2+9x-17/5
c: B(x)-A(x)=5x^4+19/20x^3-2x^2-9x+17/5
1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)
=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)
b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c) TT
a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)
=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)
=> \(\left|50x-140\right|=\left|25x+24\right|\)
=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)
Bài 2 : a. |2x - 5| = x + 1
TH1 : 2x - 5 = x + 1
=> 2x - 5 - x = 1
=> 2x - x - 5 = 1
=> 2x - x = 6
=> x = 6
TH2 : -2x + 5 = x + 1
=> -2x + 5 - x = 1
=> -2x - x + 5 = 1
=> -3x = -4
=> x = 4/3
Ba bài còn lại tương tự
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)
\(f\left(x\right)=4x^4-16x^3-11x^2+15\)
\(f\left(1\right)=4\cdot1^4-16\cdot1^3-11\cdot1^2+15\)
\(f\left(1\right)=4\cdot1^4-16\cdot1^3-11\cdot1^2+15\)
\(f\left(1\right)=-8\)
\(f\left(x\right)=4x^4-16x^3-11x^2+15\)
\(f\left(-1\right)=4\cdot\left(-1\right)^4-16\cdot\left(-1\right)^3-11\cdot\left(-1\right)^2+15\)
\(f\left(-1\right)=24\)
f(x)+g(x)=12x4-11x3+2x2-\(\frac{1}{4}\)x-\(\frac{1}{4}\)
Con f(x)-g(x) thi tru 2 da thuc tren cho nhau
