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Bài 2: 

x=13 nên x+1=14

\(f\left(x\right)=x^{14}-x^{13}\left(x+1\right)+x^{12}\left(x+1\right)-...+x^2\left(x+1\right)-x\left(x+1\right)+14\)

\(=x^{14}-x^{14}-x^{13}+x^{13}-...+x^3+x^2-x^2-x+14\)

=14-x=1

24 tháng 2 2022

x=13 nên x+1=14

f(x)=x14−x13(x+1)+x12(x+1)−...+x2(x+1)−x(x+1)+14f(x)=x14−x13(x+1)+x12(x+1)−...+x2(x+1)−x(x+1)+14

=x14−x14−x13+x13−...+x3+x2−x2−x+14=x14−x14−x13+x13−...+x3+x2−x2−x+14

=14-x=1

  
AH
Akai Haruma
Giáo viên
30 tháng 4 2023

Bài 1:
$(y+\frac{1}{3})+(y+\frac{1}{9})+(y+\frac{1}{27})+(y+\frac{1}{81})=\frac{56}{81}$

$(y+y+y+y)+(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81})=\frac{56}{81}$
$4\times y+\frac{40}{81}=\frac{56}{81}$

$4\times y=\frac{56}{81}-\frac{40}{81}=\frac{16}{81}$
$y=\frac{16}{81}:4=\frac{4}{81}$

AH
Akai Haruma
Giáo viên
30 tháng 4 2023

Bài 2:

$18: \frac{x\times 0,4+0,32}{x}+5=14$

$18: \frac{x\times 0,4+0,32}{x}=14-5=9$

$\frac{x\times 0,4+0,32}{x}=18:9=2$

$x\times 0,4+0,32=2\times x$

$2\times x-x\times 0,4=0,32$

$x\times (2-0,4)=0,32$
$x\times 1,6=0,32$
$x=0,32:1,6=0,2$

NV
28 tháng 12 2018

Câu 1/

\(f\left(13\right)=x^{13}\left(x-14\right)+14x^{12}-...-14x+14\)

\(=-x^{13}+14x^{12}-14x^{11}+...-14x+14\)

\(=x^{12}\left(-x+14\right)-14x^{11}+...-14x+14\)

\(=x^{12}-14x^{11}+...-14x+14=...\)

\(=-x+14=1\)

(Bạn để ý quy luật sau các bước rút gọn lần lượt thì mũ chẵn sẽ biến thành hệ số 1, mũ lẻ thành hệ số -1 nên x sẽ có hệ số -1)

Câu 2:

+) \(f\left(-x\right)=f\left(x\right)\) có: \(f_3\left(x\right);f_4\left(x\right);f_6\left(x\right)\)

+) \(f\left(-x\right)=-f\left(x\right)\) có: \(f_1\left(x\right);f_2\left(x\right);f_5\left(x\right)\)

+) \(f\left(x_1+x_2\right)=f\left(x_1\right)+f\left(x_2\right)\) có: \(f_1\left(x\right);f_2\left(x\right)\)

+) \(f\left(x_1x_2\right)=f\left(x_1\right).f\left(x_2\right)\) có: \(f_1\left(x\right);f_3\left(x\right);f_5\left(x\right);f_6\left(x\right)\)

7 tháng 5 2022

lm đc thì lm 0 đc thì đừng spam

7 tháng 5 2022

`( 1 + 1/12 ) xx ( 1 + 1/13 ) xx ( 1 + 1/14 ) xx ( 1 + 1/15 )`

`= 13/12 xx14/13 xx 15/14 xx 16/15`

`= ( 13 xx 14 xx 15 xx 16 )/( 12xx13xx14xx15 )`

`= 16/12`

`=4/3` 

15 tháng 7 2017

1) \(\left(-1\dfrac{1}{5}+x\right):\left(-3\dfrac{3}{5}\right)=\dfrac{-7}{4}+\dfrac{1}{4}:\dfrac{1}{8}\)

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

\(\Leftrightarrow\left(-1\dfrac{1}{5}+x\right):\left(-3\dfrac{3}{5}\right)=\dfrac{1}{4}\)

\(\Leftrightarrow-1\dfrac{1}{5}+x=\dfrac{1}{4}.\left(-3\dfrac{3}{5}\right)\)

\(\Leftrightarrow-1\dfrac{1}{5}+x=\dfrac{1}{4}.\left(-\dfrac{18}{5}\right)\)

\(\Leftrightarrow-1\dfrac{1}{5}+x=-\dfrac{9}{10}\)

\(\Leftrightarrow x=\left(-\dfrac{9}{10}\right)-\left(-1\dfrac{1}{5}\right)\)

\(\Leftrightarrow x=\dfrac{3}{10}\)

10 tháng 1 2018

2. \(x\left(x+2\right)\left(x+3\right)\left(x+5\right)=280\)

\(\Leftrightarrow x\left(x+5\right)\left(x+2\right)\left(x+3\right)=280\)

\(\Leftrightarrow\left(x^2+5x\right)\left(x^2+5x+6\right)=280\)

Đặt \(x^2+5x+3=t\)

\(\Rightarrow\left(t-3\right)\left(t+3\right)=280\)

\(\Leftrightarrow t^2-9=280\)

\(\Leftrightarrow t^2=289\Leftrightarrow\left[{}\begin{matrix}t=17\\t=-17\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+5x+3=17\\x^2+5x+3=-17\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+5x-14=0\\x^2+5x+20=0\end{matrix}\right.\)

\(\Leftrightarrow x^2+5x-14=0\text{(vì }x^2+5x+20=\left(x+\dfrac{5}{2}\right)^2+\dfrac{55}{4}>0\forall x\text{)}\)

\(\Leftrightarrow x^2-2x+7x-14=0\)

\(\Leftrightarrow x\left(x-2\right)+7\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=0\)

\(\Leftrightarrow\) x - 2 = 0 hoặc x + 7 = 0

\(\Leftrightarrow\) x = 2 hoặc x = - 7

Vậy x = 2 hoặc x = -7.

10 tháng 1 2018

3. \(\left(x+3\right)\left(x+4\right)\left(x+5\right)=x\)

\(\Leftrightarrow\left(x+3\right)\left(x+4\right)\left(x+5\right)-x=0\)

\(\Leftrightarrow x^3+12x^2+47x+60-x=0\)

\(\Leftrightarrow x^3+12x^2+46x+60=0\)

\(\Leftrightarrow x^3+6x^2+6x^2+36x+10x+60=0\)

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

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

\(\Leftrightarrow x+6=0\text{(vì }x^2+6x+10=\left(x+3\right)^2+1>0\forall x\text{)}\)

\(\Leftrightarrow x=-6\)

Vậy x = -6.

11 tháng 7 2017

\(a.-8:\left(4\dfrac{1}{5}x+\dfrac{3}{10}\right)=4\dfrac{4}{9}\)

\(4\dfrac{1}{5}x+\dfrac{3}{10}=\left(-8\right):4\dfrac{4}{9}\)

\(4\dfrac{1}{5}x+\dfrac{3}{10}=\dfrac{-9}{5}\)

\(4\dfrac{1}{5}x=\dfrac{-9}{5}-\dfrac{3}{10}\)

\(4\dfrac{1}{5}x=\dfrac{-21}{10}\)

\(x=\dfrac{-21}{10}:\dfrac{21}{5}\)

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

Vay \(x=\dfrac{-1}{2}\).

\(b.4\dfrac{2}{3}-\left(\dfrac{3}{5}:x\right)=-20\%\)

\(\dfrac{14}{3}-\left(\dfrac{3}{5}:x\right)=\dfrac{-1}{5}\)

\(\dfrac{3}{5}:x=\dfrac{14}{3}-\dfrac{-1}{5}\)

\(\dfrac{3}{5}:x=\dfrac{73}{15}\)

\(x=\dfrac{3}{5}:\dfrac{73}{15}\)

\(x=\dfrac{9}{73}\)

Vay \(x=\dfrac{9}{73}\).

Câu c; d; e tương tự nhé.

d: =>-x-5/6=7/12-4/12=3/12=1/4

=>-x=1/4+5/6=13/12

hay x=-13/12

e: =>x+3=-5

hay x=-8

f: =>4,5-2x=-1/2

=>2x=5

hay x=5/2

8 tháng 3 2022

nhanh v ???

20 tháng 12 2022

\(\dfrac{15}{12}+\dfrac{5}{13}-\dfrac{3}{12}-\dfrac{18}{13}-\dfrac{1}{3}\)

\(=\left(\dfrac{15}{12}-\dfrac{3}{12}\right)+\left(\dfrac{5}{13}-\dfrac{18}{13}\right)-\dfrac{1}{3}\)

\(=-1+1-\dfrac{1}{3}\)

\(=0-\dfrac{1}{3}\)

\(=\dfrac{-1}{3}\)

------------------------------------------

\(14.\dfrac{3}{2}+\dfrac{6}{5}:\left(-\dfrac{2}{5}\right)\)

\(=14.\dfrac{3}{2}+\dfrac{6}{5}.\dfrac{-5}{2}\)

\(=21+\dfrac{6}{5}.\dfrac{-5}{2}\)

\(=21+\left(-3\right)\)

\(=18\)

------------------------------------------------

\(\sqrt{\dfrac{1}{4}+\dfrac{2}{3}-\left(\dfrac{1}{3}\right)^2}\)

\(=\sqrt{\dfrac{1}{4}+\dfrac{2}{3}-\dfrac{1}{9}}\)

\(=\sqrt{\dfrac{3}{12}+\dfrac{8}{12}-\dfrac{1}{9}}\)

\(=\sqrt{\dfrac{11}{12}-\dfrac{1}{9}}\)

\(=\sqrt{\dfrac{99}{108}-\dfrac{12}{108}}\)

\(=\sqrt{\dfrac{29}{36}}\)

\(=\dfrac{\sqrt{29}}{6}\)

20 tháng 12 2022

\(\dfrac{15}{12}+\dfrac{5}{13}-\dfrac{3}{12}-\dfrac{18}{13}-\dfrac{1}{3}\)
\(=\dfrac{5}{4}+\dfrac{5}{13}-\dfrac{1}{4}-\dfrac{18}{13}-\dfrac{1}{3}\)
\(=\left(\dfrac{5}{4}-\dfrac{1}{4}\right)+\left(\dfrac{5}{13}-\dfrac{18}{13}\right)-\dfrac{1}{3}\)
\(=1+\left(-1\right)-\dfrac{1}{3}=0-\dfrac{1}{3}=-\dfrac{1}{3}\)