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

=>8x^2+4x-1-5=0

=>8x^2+4x-6=0

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

=>\(x=\dfrac{-1\pm\sqrt{13}}{4}\)

1: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5x+2\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(-4x+1\right)=0\)

hay \(x\in\left\{3;\dfrac{1}{4}\right\}\)

2: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)-\left(x-1\right)\left(x^2-2x+16\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1-x^2+2x-16\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(3x-15\right)=0\)

hay \(x\in\left\{1;5\right\}\)

3: \(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x-1\right)\left(2x+1\right)=0\)

hay \(x\in\left\{1;\dfrac{1}{2};-\dfrac{1}{2}\right\}\)

4: \(\Leftrightarrow x^2\left(x+4\right)-9\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x-3\right)\left(x+3\right)=0\)

hay \(x\in\left\{-4;3;-3\right\}\)

5: \(\Leftrightarrow\left[{}\begin{matrix}3x+5=x-1\\3x+5=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-6\\4x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)

6: \(\Leftrightarrow\left(6x+3\right)^2-\left(2x-10\right)^2=0\)

\(\Leftrightarrow\left(6x+3-2x+10\right)\left(6x+3+2x-10\right)=0\)

\(\Leftrightarrow\left(4x+13\right)\left(8x-7\right)=0\)

hay \(x\in\left\{-\dfrac{13}{4};\dfrac{7}{8}\right\}\)

1.

\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=\left(x-3\right)\left(5x-2\right)\)

\(\Leftrightarrow x+3=5x-2\)

\(\Leftrightarrow4x=5\Leftrightarrow x=\dfrac{5}{4}\)

2.

\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=\left(x-1\right)\left(x^2-2x+16\right)\)

\(\Leftrightarrow x^2+x+1=x^2-2x+16\)

\(\Leftrightarrow3x=15\Leftrightarrow x=5\)

3.

\(\Leftrightarrow4x^2\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2};x=-\dfrac{1}{2}\end{matrix}\right.\)

7.

\(\Leftrightarrow x^2+2x-15=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

8.\(\Leftrightarrow x^4+x^3+4x^3+4x^2=0\)

\(\Leftrightarrow x^3\left(x+1\right)+4x^2\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^3+4x^2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=0;x=-4\end{matrix}\right.\)

9.\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=\left(x-2\right)\left(3-2x\right)\)

\(\Leftrightarrow x+2=3-2x\)

\(\Leftrightarrow3x=1\Leftrightarrow x=\dfrac{1}{3}\)

a) C1 : 2/5 x 3/7 + 2/7 x 4/7

         = 2/5 x (3/7 + 4/7)

        = 2/5 x 1

        = 2/5

C2 : 2/5 x 3/7 + 2/5 x 4/7

= 6/35 + 8/35

= 2/5 

b) (2/3 - 4/7) x 5/5

= 2/21 x 5/5

= 2/21

C2 : (2/3 - 4/7) x 5/5

= 2/3 x 1 - 4/7 x 1

= 2/3 - 4/7

= 2/21

c) 3/4 x 2/5 - 3/4 x 2/7

= 3/4 x (2/5 - 2/7)

= 3/4 x 4/35

= 3/35

C2 : 3/4 x 2/5 - 3/4 x 2/7

= 3/10 - 3/14

= 3/35

a, 

Cách 1

\(\frac{2}{5}\times\frac{3}{7}+\frac{2}{5}\times\frac{4}{7}\)

\(=\frac{6}{35}+\frac{8}{35}\)

\(=\frac{2}{5}\)

Cách 2 :

\(\frac{2}{5}\times\frac{3}{7}+\frac{2}{5}\times\frac{4}{7}\)

\(=\frac{2}{5}\times\left(\frac{3}{7}+\frac{4}{7}\right)\)

\(=\frac{2}{5}\times1\)

\(=\frac{2}{5}\)

b,

Cách 1 :

\(\left(\frac{2}{3}-\frac{4}{7}\right)\times\frac{5}{5}\)

\(=\frac{2}{21}\times1\)

\(=\frac{2}{21}\)

Cách 2 :

\(\left(\frac{2}{3}-\frac{4}{7}\right)\times\frac{5}{5}\)

\(=\frac{2}{3}\times\frac{5}{5}-\frac{4}{7}\times\frac{5}{5}\)

\(=\frac{2}{3}-\frac{4}{7}\)

\(=\frac{2}{21}\)

c, 

Cách 1 :

\(\frac{3}{4}\times\frac{2}{5}-\frac{3}{4}\times\frac{2}{7}\)

\(=\frac{3}{10}-\frac{3}{14}\)

\(=\frac{3}{35}\)

Cách 2 :

\(\frac{3}{4}\times\frac{2}{5}-\frac{3}{4}\times\frac{2}{7}\)

\(=\frac{3}{4}\times\left(\frac{2}{5}-\frac{2}{7}\right)\)

\(=\frac{3}{4}\times\frac{4}{35}\)

\(=\frac{3}{35}\)

Mk nhanh nhất đó

Đúng 100%

Tk mk mk tk lại

Cảm ơn bạn nhiều

Thank you very  much

( ^ _ ^ )

\(A=5x^3-7x^2+3x^3-4x^2+x^2-x^3+5x-1=7x^3-10x^2+5x-1\)

\(B=5x^3+3x^2-7x^4-5x^3+4x^2-x^4+3=-8x^4+7x^2+3\)

\(A=7x^3-10x^2+5x-1\)

\(B=-8x^4+7x^2+3\)

\(a,\frac{2}{5}:\frac{6}{5}+\frac{3}{4}.\frac{12}{5}\)

=\(\frac{2}{5}.\frac{5}{6}+\frac{9}{5}\)

=\(\frac{1}{3}+\frac{9}{5}\)

=\(\frac{32}{5}\)

b)  \(\frac{17}{5}.\frac{3}{5}.\frac{17}{5}.\frac{2}{5}\)

=\(\frac{17}{5}.\left(\frac{3}{5}.\frac{2}{5}\right)\)

=\(\frac{17}{5}.\frac{6}{25}\)

=\(\frac{102}{125}\)

c)\(\frac{1}{4}+\frac{3}{4}.\frac{2}{3}-\frac{1}{2}\)

=\(\frac{1}{4}+\frac{1}{2}-\frac{1}{2}\)

=\(\frac{1}{4}\)

d)\(\frac{6}{7}+\frac{5}{8}:5\)

=\(\frac{6}{7}+\frac{5}{8}.\frac{1}{5}\)

=\(\frac{6}{7}+\frac{1}{8}\)

=\(\frac{48}{56}+\frac{7}{56}\)

=\(\frac{55}{56}\)

tk nhé bn!

a) \(2x\left(x^2-7x-3\right)=2x.x^2-2x.7x-2x.3=2x^3-14x^2-6x\)

b) \(\left(-2x^3+y^2-7xy\right)4xy^2=\left(-2x^3\right)4xy^2+y^24xy^2-7xy.4xy^2=-8x^4y^2+4xy^4-28x^2y^3\)

c) \(\left(-5x^3\right)\left(2x^2+3x-5\right)=-5x^32x^2-5x^33x-5x^3.-5=-10x^5-15x^4+25x^3\)

d) \(\left(2x^2-xy+y^2\right)\left(-3x^3\right)=-3x^32x^2-3x^3.-xy-3x^3y^2=-6x^5+3x^4y-3x^3y^2\)

e) \(\left(x^2-2x+3\right)\left(x-4\right)=x\left(x^2-2x+3\right)-4\left(x^2-2x+3\right)=x^3-2x^2+3x-4x^2+8x-12=x^3-6x^2+11x-12\)

f) \(\left(2x^3-3x-1\right)\left(5x+2\right)=5x\left(2x^3-3x-1\right)+2\left(2x^3-3x-1\right)=10x^4-15x^2-5x+4x^3-6x-2=10x^4+4x^3-15x^2-11x-2\)

g)

\(\left(25x^2+10xy+4y^2\right).\left((5x-2y\right)\)

\(=125x^3-50x^2y+20x^2y-20xy^2+20xy^2-8y^3\)

\(=125x^3-30x^2y+8y^3\)

h)

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

\(=20x^5-5x^4+10x^3-4x^4+x^3-2x^2+8x^3-2x^2+4x-12x^2+3x-6\)

\(=20x^5-9x^4+19x^3-16x^2+7x-6\)

Ta có:

Chọn đáp án B