(x+1)(x+3)(x+5)(x+7) + 15

= [ (x+1)(x+7) ].[ (x+3)(x+5) ] + 15

= (x² + 7x + x + 7).(x² + 5x + 3x + 15) + 15

= (x² + 8x + 7).(x² + 8x + 15) + 15

= (x² + 8x + 11 - 4)(x² + 8x + 11 + 4) + 15.

Đặt x² + 8x + 11 = y (1) ta được :

(t - 4)(t + 4) + 15 = t² - 16 + 15 = t² - 1 = (t+1)(t-1) (2).

Thay (1) vào (2) ta được: đa thức trên được phân tích thành:

(x² + 8x + 11 + 1)(x² + 8x + 11 - 1)

= (x² + 8x + 12)(x² + 8x + 10).

Chúc bn học tốt!

\(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)

\(=x^4+9x^3+23x^2+15x+7x^3+63x^2+161x+105+15\)

\(=x^4+16x^3+86x^2+176x+120\)

\(=\left(x+2\right)\left(x+6\right)\left(x^2+8x+10\right)\)

\(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)

=\(\left(x+1\right)\left(x+7\right)\left(x+3\right)\left(x+5\right)+15\)

=\(\left(x^2+8x+7\right)\)\(\left(x^2+8x+15\right)+15\) (1)

Đặt \(x^2+8x+11=a\) thay vảo (1) đc:

\(\left(a-4\right)\left(a+4\right)+15=a^2-16+15=a^2-1\)=\(\left(a-1\right)\left(a+1\right)\)=\((x^2+8x+10)\left(x^2+8x+12\right)\)

các bn lm đến đâu cx dc miễn là lm hộ mk cái ạ, ai đang lm vào nhắn tin vs mk để mk bít nha

a; \(-\dfrac{8}{3}+\dfrac{7}{5}-\dfrac{71}{15}< x< -\dfrac{13}{7}+\dfrac{19}{14}-\dfrac{7}{2}\)

              -\(\dfrac{19}{15}\) - \(\dfrac{71}{15}\) < \(x\) < -\(\dfrac{1}{2}\) - \(\dfrac{7}{2}\)

              -6 < \(x\) < -4

             vì \(x\) \(\in\) Z nên \(x\) = -5

- \(\dfrac{19}{6}+\dfrac{-15}{2}+\dfrac{11}{3}\) < \(x\) < - \(\dfrac{5}{4}\) + \(\dfrac{19}{12}\) - \(\dfrac{10}{3}\)

- \(\dfrac{32}{3}+\dfrac{11}{3}< x< \) \(\dfrac{1}{3}\) - \(\dfrac{10}{3}\)

-7 < \(x\) < - 3

Vì \(x\in\) Z nên \(x\in\) {-6;-5;-4}

\(a,\frac{62}{7}:x=\frac{29}{9}:\frac{3}{56}\)

\(\frac{62}{7}:x=\frac{1624}{27}\)

\(x=\frac{62}{7}:\frac{1624}{27}=\frac{837}{5684}\)

\(b,\frac{1}{5}:x=\frac{1}{5}-\frac{1}{7}\)

\(\frac{1}{5}:x=\frac{2}{35}\)

\(x=\frac{1}{5}:\frac{2}{35}=\frac{7}{2}\)

\(c,\frac{2}{3}.x-\frac{4}{7}=\frac{1}{7}\)

\(\frac{2}{3}.x=\frac{1}{7}+\frac{4}{7}=\frac{5}{7}\)

\(x=\frac{5}{7}:\frac{2}{3}=\frac{15}{14}\)

\(d,\frac{2}{7}-\frac{8}{9}.x=\frac{2}{3}\)

\(\frac{8}{9}.x=\frac{2}{7}-\frac{2}{3}=-\frac{8}{21}\)

\(x=-\frac{8}{21}:\frac{8}{9}=-\frac{3}{7}\)

\(e,\frac{4}{7}+\frac{5}{9}:x=\frac{1}{5}\)

\(\frac{5}{9}:x=\frac{1}{5}-\frac{4}{7}=-\frac{13}{35}\)

\(x=\frac{5}{9}:-\frac{13}{35}=\frac{175}{117}\)

\(i,\frac{2}{5}-\frac{2}{5}.x=\frac{2}{5}\)

\(\frac{2}{5}.\left(1-x\right)=\frac{2}{5}\)

\(1-x=\frac{2}{5}:\frac{2}{5}=1\)

\(x=1-1=0\)

\(g,\frac{2}{3}+\frac{1}{3}:x=-1\)

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

\(x=\frac{1}{3}:-\frac{5}{3}=-\frac{1}{5}\)

học tốt nha

Ta có : 5x³ - 45x

= 5x(x² - 9)

= 5x(x - 3)(x + 3)

b ) Ta có : 3x² - 7x - 6 

= 3x² - 9x + 2x - 6 

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

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

a. \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{x-2\sqrt{x}}\right)\cdot\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{4}{x-4}\right)\)

<=> \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{\sqrt{x}-2+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

<=> \(P=\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

<=> \(P=\dfrac{\sqrt{x}+2}{x-2\sqrt{x}}\)

b. Khi \(x=7+4\sqrt{3}=\left(2+\sqrt{3}\right)^2\) => \(\sqrt{x}=2+\sqrt{3}\)

=> \(P=\dfrac{2+\sqrt{3}+2}{7+4\sqrt{3}-2\left(2+\sqrt{3}\right)}=\dfrac{4+\sqrt{3}}{7+4\sqrt{3}-4-2\sqrt{3}}=\dfrac{4+\sqrt{3}}{3+2\sqrt{3}}=\dfrac{5\sqrt{3}-6}{3}\)

check giùm mik

a: Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{x-2\sqrt{x}}\right)\cdot\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{4}{x-4}\right)\)

\(=\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}-2+4}{x-4}\)

\(=\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

b: Thay \(x=7+4\sqrt{3}\) vào P, ta được:

\(P=\dfrac{2+\sqrt{3}+2}{\sqrt{3}\left(2+\sqrt{3}\right)}=\dfrac{-6+5\sqrt{3}}{3}\)

1,x=3 hoặc x=-2

2,x=12

3,không có x nào thỏa mãn

Bài 1 : 

Ta có : 

\(\left|2x-1\right|=5\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}2x-1=5\\2x-1=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=6\\2x=-4\end{cases}}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{6}{2}\\x=\frac{-4}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}}\)

Vậy \(x=-2\) hoặc \(x=3\)

Bài 2 : 

Đặt \(A=\frac{3x+4}{x-1}\) ta có : 

\(A=\frac{3x+4}{x-1}=\frac{3x-3+7}{x-1}=\frac{3x-3}{x-1}+\frac{7}{x-1}=\frac{3\left(x-1\right)}{x-1}+\frac{7}{x-1}=3+\frac{7}{x-1}\)

Để A là số nguyên thì \(\frac{7}{x-1}\) phải nguyên \(\Rightarrow\)\(7⋮\left(x-1\right)\)\(\Rightarrow\)\(\left(x-1\right)\inƯ\left(7\right)\)

Mà \(Ư\left(7\right)=\left\{1;-1;7;-7\right\}\)

Suy ra : 

Vậy \(x\in\left\{-6;0;2;8\right\}\) thì \(A\inℤ\)

Chúc bạn học tốt ~ 

cho câu lớp 3 đi mà

a)   \(x^3-5x^2+8x-4\)

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

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

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

a)   \(x^3-5x^2+8x-4\)

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

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

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