7x2+7x3=???
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
7x3+7x2+7x3+7
= 7x (3+2+3+7)
= 7x 15
= 105
![](https://rs.olm.vn/images/avt/0.png?1311)
\(Sửa:\left(2x^4-7x^3-7x^2-6x-2\right):\left(2x^2+x-1\right)\\ =\left(2x^4+x^3-x^2-8x^3-4x^2+4x-2x^2-x+1-9x-3\right):\left(2x^2+x-1\right)\\ =\left[x^2\left(2x^2+x-1\right)-4x\left(2x^2+x-1\right)-\left(2x^2+x-1\right)-9x-3\right]:\left(2x^2+x-1\right)\\ =x^2-4x-1\left(\text{dư }-9x-3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$A(x)+B(x)=5-3x^2+2x-5x^3+6x+7x^3-7x^2-9$
$=2x^3-10x^2+8x-4$
$A(x)-B(x)=(5-3x^2+2x-5x^3)-(6x+7x^3-7x^2-9)$
$=-12x^3+4x^2-4x+14$
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
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 ) \(\frac{2}{7}\cdot\frac{3}{9}+\frac{2}{7}\cdot\frac{2}{3}=\frac{2}{7}\cdot\left(\frac{3}{9}+\frac{2}{3}\right)=\frac{2}{7}\cdot1=\frac{2}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(-\frac{23}{7}.\frac{3}{10}+\frac{13}{7}.\frac{3}{10}\)
\(=\frac{3}{10}.\left(\left(-\frac{23}{7}\right)+\frac{13}{7}\right)\)
\(=\frac{3}{10}.\left(\frac{-10}{7}\right)\)
\(=\frac{-3}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(A\left(x\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Chọn B
Ta có: B(x) = 6x4 - 7x3 + 6x2- 7x3 + 4x4 + 3 - 5x + 2x
= 10x4 - 14x3 + 6x2 - 3x + 3.
35
1100