\(Sửa:\left(2x^4-7x^3-7x^2-6x-2\right):\left(2x^2+x-1\right)\\ =\left(2x^4+x^3-x^2-8x^3-4x^2+4x-2x^2-x+1-9x-3\right):\left(2x^2+x-1\right)\\ =\left[x^2\left(2x^2+x-1\right)-4x\left(2x^2+x-1\right)-\left(2x^2+x-1\right)-9x-3\right]:\left(2x^2+x-1\right)\\ =x^2-4x-1\left(\text{dư }-9x-3\right)\)

dễ lắm

3/7\(\times\)(2/5+3/5+1)

=3/7 nhân 2

=6/7

Lời giải:
$A(x)+B(x)=5-3x^2+2x-5x^3+6x+7x^3-7x^2-9$

$=2x^3-10x^2+8x-4$

$A(x)-B(x)=(5-3x^2+2x-5x^3)-(6x+7x^3-7x^2-9)$

$=-12x^3+4x^2-4x+14$

7x1=7

7x2=14

7x3=21

7x4=28

7x5=40

7x6=42

7x7=49

7x8=56

7x9=63

7x10=70

7x1=7

7x2=14

7x3=21

7x4=28

7x5=35

7x6=42

7x7=49

7x8=56

7x9=63

7x10=70

7×1=7

7×2=14

7×3=21

7×4=28

7×5=35

7×6=42

7×7=49

7×8=56

7×9=63

7×10=70

<tích cho mk nhé>

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a) 563 x 23 + 23 x 28 - 23

= 23 x (563 + 28 - 1)

= 23 x 600

= 19200

b) \(\frac{7}{11}+\frac{3}{4}+\frac{4}{11}+\frac{1}{4}-1\frac{2}{5}=\left(\frac{7}{11}+\frac{4}{11}\right)+\left(\frac{3}{4}+\frac{1}{4}\right)-\frac{7}{5}=2-\frac{7}{5}=\frac{3}{5}\)

c) \(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times...\times\left(1-\frac{1}{10}\right)=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times....\times\frac{9}{10}\)

\(=\frac{1\times2\times3\times...\times9}{2\times3\times4\times...\times10}=\frac{1}{10}\)

d) \(\frac{2}{7}\times\frac{3}{9}+\frac{2}{7}\times\frac{2}{3}=\frac{2}{7}\times\left(\frac{3}{9}+\frac{2}{3}\right)=\frac{2}{7}\times1=\frac{2}{7}\)

a ) \(563\cdot23+23\cdot28-23\)

\(=23\cdot\left(563+28-1\right)\)

\(=23\cdot600\)

\(=19200\)

b ) \(\frac{7}{11}+\frac{3}{4}+\frac{4}{11}+\frac{1}{4}-1\frac{2}{5}=\left(\frac{7}{11}+\frac{4}{11}\right)+\left(\frac{3}{4}+\frac{1}{4}\right)-\frac{7}{5}=2-\frac{7}{5}=\frac{3}{5}\)

c ) \(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{10}\right)=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{9}{10}\)

\(=\frac{1\cdot2\cdot3\cdot...\cdot9}{2\cdot3\cdot4\cdot...\cdot10}=\frac{1}{10}\)

d ) \(\frac{2}{7}\cdot\frac{3}{9}+\frac{2}{7}\cdot\frac{2}{3}=\frac{2}{7}\cdot\left(\frac{3}{9}+\frac{2}{3}\right)=\frac{2}{7}\cdot1=\frac{2}{7}\)

54/35

\(-\frac{23}{7}.\frac{3}{10}+\frac{13}{7}.\frac{3}{10}\)

\(=\frac{3}{10}.\left(\left(-\frac{23}{7}\right)+\frac{13}{7}\right)\)

\(=\frac{3}{10}.\left(\frac{-10}{7}\right)\)

\(=\frac{-3}{7}\)

`A(x)=7x^3-5x^2-7x+3-7x^3+5x^2+17x+27`

`A(x)=(7x^3-7x^3)-(5x^2-5x^2)+(-7x+17x)+(3+27)`

`A(x)=10x+30`

Cho `A(x)=0`

`=>10x+30=0`

`=>10x=-30`

`=>x=-3`

Vậy nghiệm của đa thức `A(x)` là `x=-3`

Đặt \(A\left(x\right)=0\)

\(\rightarrow7x^3-5x^2-7x+3-7x^3+5x^2+17x+27=0\)

\(\Leftrightarrow10x+30=0\)

\(\Leftrightarrow10x=-30\)

\(\Leftrightarrow x=-3\)

Vậy \(x=-3\) là nghiệm của đa thức \(A\left(x\right)\)

A(x) = 7x³ - 5x² - 7x + 3 - 7x³ + 5x² +17x +27

= 10x + 30

đặt A(x) = 0

<=> 10x + 30 = 0

<=> 10x = -30

<=> x = -3

vậy x = -3 là nghiệm của A(x)