\(4,7\div0,25+5,3\times4\)

\(=18,8+21,2\)

\(=40\)

\(3\times\left(a-2\right)+150=240\)

\(3\times\left(a-2\right)=90\)

\(a-2=30\)

\(a=32\)

\(\dfrac{1}{9}+a+\dfrac{7}{12}=\dfrac{17}{18}\)

\(\dfrac{1}{9}+a=\dfrac{13}{36}\)

\(a=\dfrac{1}{4}\)

\(\left(\dfrac{1}{2}\times\dfrac{1}{3}+\dfrac{1}{3}\times\dfrac{1}{4}+\dfrac{1}{4}\times\dfrac{1}{5}+\dfrac{1}{5}\times\dfrac{1}{6}+\dfrac{1}{6}\times\dfrac{1}{7}+\dfrac{1}{7}\times\dfrac{1}{8}\right)\times a=\dfrac{9}{16}\)

\(\left(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+\dfrac{1}{7\times8}\right)\times a=\dfrac{9}{16}\)

\(\left(\dfrac{1}{2}-\dfrac{1}{8}\right)\times a=\dfrac{9}{16}\)

\(\dfrac{3}{8}\times a=\dfrac{9}{16}\)

\(a=\dfrac{3}{2}\)

a: P(x)=5x^2-4x+7

Sửa đề: Q(x)=-5x^3-x^2+4x-5

Q(x)+P(x)+5x^2-2=0

=>5x^2-4x+7-5x^3-x^2+4x-5+5x^2-2=0

=>5x^3=0

=>x=0

+) \(\dfrac{1}{3}x=-\dfrac{4}{3}-\dfrac{1}{2}=-\dfrac{11}{6}\)

\(x=-\dfrac{11}{6}:\dfrac{1}{3}=-\dfrac{11}{2}\)

+) \(\dfrac{4}{3}x=-\dfrac{2}{3}+\dfrac{1}{2}=-\dfrac{1}{6}\)

\(x=-\dfrac{1}{6}:\dfrac{4}{3}=-\dfrac{1}{8}\)

+) \(2\left(x-1\right)=\dfrac{5}{2}+\dfrac{2}{3}=\dfrac{19}{6}\)

\(x-1=\dfrac{19}{12}\)

\(x=\dfrac{31}{12}\)

\(\dfrac{1}{3}x+\dfrac{1}{2}=-\dfrac{4}{3}\)

\(\dfrac{1}{3}x=\left(-\dfrac{4}{3}\right)-\dfrac{1}{2}\)

\(\dfrac{1}{3}x=-\dfrac{11}{6}\)

\(x=\left(-\dfrac{11}{6}\right):\dfrac{1}{3}\)

\(x=-\dfrac{11}{2}\)

\(-\dfrac{2}{3}-\dfrac{4}{3}x=-\dfrac{1}{2}\)

\(\dfrac{4}{3}x=\left(-\dfrac{2}{3}\right)-\dfrac{-1}{2}\)

\(\dfrac{4}{3}x=-\dfrac{1}{6}\)

\(x=\left(-\dfrac{1}{6}\right):\dfrac{4}{3}\)

\(x=-\dfrac{1}{8}\)

\(\dfrac{5}{2}-2\left(x-1\right)=-\dfrac{2}{3}\)

\(2\left(x-1\right)=\dfrac{5}{2}-\left(-\dfrac{2}{3}\right)\)

\(2\left(x-1\right)=\dfrac{19}{6}\)

\(\left(x-1\right)=\dfrac{19}{6}:2\)

\(x-1=\dfrac{19}{12}\)

\(x=\dfrac{19}{12}+1\)

\(x=\dfrac{31}{12}\)

\(\dfrac{7}{-2}x-\dfrac{1}{3}=-\dfrac{1}{2}\)

\(\dfrac{-7}{2}x-\dfrac{1}{3}=-\dfrac{1}{2}\)

\(\dfrac{-7}{2}x=\left(-\dfrac{1}{2}\right)+\dfrac{1}{3}\)

\(\dfrac{-7}{2}x=-\dfrac{1}{6}\)

\(x=\left(-\dfrac{1}{6}\right):\left(-\dfrac{7}{2}\right)\)

\(x=\dfrac{1}{21}\)

\(\dfrac{8}{5}-\dfrac{1}{2}:x=\dfrac{8}{3}\)

\(\dfrac{1}{2}:x=\dfrac{8}{5}-\dfrac{8}{3}\)

\(\dfrac{1}{2}:x=-\dfrac{16}{15}\)

\(x=\dfrac{1}{2}:\left(-\dfrac{16}{15}\right)\)

\(x=-\dfrac{15}{32}\)

\(-\dfrac{5}{4}x-\dfrac{1}{2}x=-\dfrac{7}{3}\)

\(x\cdot\left(-\dfrac{5}{4}-\dfrac{1}{2}\right)=-\dfrac{7}{3}\)

\(x\cdot\left(-\dfrac{7}{4}\right)=-\dfrac{7}{3}\)

\(x=\left(-\dfrac{7}{3}\right):\left(-\dfrac{7}{4}\right)\)

\(x=\dfrac{4}{3}\)

Để thu gọn và sắp xếp các hạng tử của mỗi đa thức, ta cần thực hiện các bước sau:
Đối với đa thức P(x): P(x) = (4x + 1 - x^2 + 2x^3) - (x^4 + 3x - x^3 - 2x^2 - 5) = 4x + 1 - x^2 + 2x^3 - x^4 - 3x + x^3 + 2x^2 + 5 = -x^4 + 3x^3 + x^2 + x + 6
Đối với đa thức Q(x): Q(x) = 3x^4 + 2x^5 - 3x - 5x^4 - x^5 + x + 2x^5 - 1 = 2x^5 - x^5 + 3x^4 - 5x^4 + x - 3x - 1 = x^5 - 2x^4 - 2x - 1
Sau khi thu gọn và sắp xếp các hạng tử, ta có: P(x) = -x^4 + 3x^3 + x^2 + x + 6 Q(x) = x^5 - 2x^4 - 2x - 1

a: \(P\left(x\right)=\left(4x+1-x^2+2x^3\right)-\left(x^4+3x-x^3-2x^2-5\right)\)

\(=4x+1-x^2+2x^3-x^4-3x+x^3+2x^2+5\)

\(=-x^4+3x^3+x^2+x+6\)

\(Q\left(x\right)=3x^4+2x^5-3x-5x^4-x^5+x+2x^5-1\)

\(=\left(2x^5-x^5+2x^5\right)+\left(3x^4-5x^4\right)+\left(-3x+x\right)-1\)

\(=-x^5-2x^4-2x-1\)

b: Bạn ghi lại đề đi bạn

`C(x)=`\(5-8x^4+2x^3+x+5x^4+x^2-4x^3\)

`C(x)= (-8x^4+5x^4)+(2x^3-4x^3)+x^2+x+5`

`C(x)= -3x^4-2x^3+x^2+x+5`

`D(x)=`\(\left(3x^5+x^4-4x\right)-\left(4x^3-7+2x^4+3x^5\right)\)

`D(x)= 3x^5+x^4-4x-4x^3+7-2x^4-3x^5`

`D(x)=(3x^5-3x^5)+(x^4-2x^4)-4x^3-4x+7`

`D(x)=-x^4-4x^3-4x+7`

`P(x)=C(x)+D(x)`

`P(x)=( -3x^4-2x^3+x^2+x+5)+(-x^4-4x^3-4x+7)`

`P(x)=-3x^4-2x^3+x^2+x+5-x^4-4x^3-4x+7`

`P(x)=(-3x^4-x^4)+(-2x^3-4x^3)+x^2+(x-4x)+(5+7)`

`P(x)=-4x^4-6x^3+x^2-3x+12`

`Q(x)=C(x)-D(x)`

`Q(x)=( -3x^4-2x^3+x^2+x+5)-(-x^4-4x^3-4x+7)`

`Q(x)=-3x^4-2x^3+x^2+x+5+x^4+4x^3+4x-7`

`Q(x)=(-3x^4+x^4)+(-2x^3+4x^3)+x^2+(x+4x)+(5-7)`

`Q(x)=-2x^4+2x^3+x^2+5x-2`

`F(x)=Q(x)-(-2x^4+2x^3+x^2-12)`

`F(x)=(-2x^4+2x^3+x^2+5x-2)-(-2x^4+2x^3+x^2-12)`

`F(x)=-2x^4+2x^3+x^2+5x-2+2x^4-2x^3-x^2+12`

`F(x)=(-2x^4+2x^4)+(2x^3-2x^3)+(x^2-x^2)+5x+(-2+12)`

`F(x)=5x+10`

Đặt `5x+10=0`

`\Leftrightarrow 5x=0-10`

`\Leftrightarrow 5x=-10`

`\Leftrightarrow x=-10 \div 5`

`\Leftrightarrow x=-2`

Vậy, nghiệm của đa thức là `x=-2.`

\(P\left(x\right)-Q\left(x\right)\)

\(=x^3+3x^3-x-4-x^4-3x^3+3x^5+2x-6\)

\(=3x^5+x-10\)

\(P\left(x\right)-Q\left(x\right)=3x^2-5+x^4-x+1-6+2x-3x^3-x^4+3x^5\\ =3x^5-3x^3+3x^2+x-10\\ Q\left(x\right)-P\left(x\right)=6-2x+3x^3+x^4-3x^5-3x^2+5-x^4+x-1=-3x^5+3x^3-3x^2-x+10\)

Đó là 2 biểu thức đối nhau

Các hệ số của 2 đa thức đối nhau

Ta có :

\(B\left(x\right)=3x^4+x^5-2\left(x^3+4\right)-10x^2+9x\)

\(=x^5+3x^4-2x^3-10x^2+9x-8\)

\(C\left(x\right)=A\left(x\right)-B\left(x\right)\)

\(=x^5+3x^4-2x^3-9x^2+11x-6-\left(x^5+3x^4-2x^3-10x^2+9x-8\right)\)

\(=x^5+3x^4-2x^3-9x^2+11x-6-x^5-3x^4+2x^3+10x^2-9x+8\)

\(=x^2-2x+2\)

ai đó làm giúp mik

cảm ơn

dễ thé mà ko biet lam

Ko có rút gọn sao mà làm