bn hãy lập bảng ra thì lm đc

\(a,x=1\)

\(b,x=3\)

\(c,x=0\)

\(d,x=0,1,2,3,4,....\)

\(e,x=2,4\)

a)x=1

b)x=3

c)x=0

d)0,1,2,3...

e)x=2,4

ai k mình mình k lại cho

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

vì \(\left(2x+3\right)^2\ge0;\left(3x-2\right)^4\ge0\)

nên\(\Rightarrow\hept{\begin{cases}\left(2x+3\right)^2=0\\\left(3x-2\right)^4=0\end{cases}\Rightarrow\hept{\begin{cases}2x+3=0\\3x-2=0\end{cases}}}\)

\(\Rightarrow\hept{\begin{cases}x=-\frac{3}{2}\\x=\frac{2}{3}\end{cases}}\)

Bạn làm như trên \(\uparrow\)sau đó thì kết luận :

Vậy không có giá trị x nào thỏa mản (2x + 3)² + (3x - 2)⁴ = 0 .

Bài nào vậy???

bài nào vậy bn

0 - 1 + 2 - 3 + 4 - 5 + 6 - 7 + 8 = 4

bằng 4

giúp mình đi , mình đang cần gấp. huhu

3h 44 phút

12:

\(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{5}\right)...\left(1-\dfrac{1}{2004}\right)\)

\(B=\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\left(\dfrac{3}{3}-\dfrac{1}{3}\right)\left(\dfrac{4}{4}-\dfrac{1}{4}\right)...\left(\dfrac{2004}{2004}-\dfrac{1}{2004}\right)\)

\(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}...\cdot\dfrac{2003}{2004}\)

\(B=\dfrac{1\cdot2\cdot3\cdot...\cdot2002\cdot2003}{2\cdot3\cdot4\cdot...\cdot2003\cdot2004}\)

\(B=\dfrac{1}{2004}\)

`12) B=(1-1/2)(1-1/3)(1-1/4)(1-1/5)....(1-1/2003)(1-1/2004)`

    `B=1/2 . 2/3 . 3/4 . 4/5 .... 2002/2003 . 2003/2004`

    `B= [1.2.3.4......2002.2003]/[2.3.4.5.....2003.2004]`

    `B=1/2004`

`7)`

`a)[254xx399-145]/[254+399xx253]`

`=[(253+1)xx399-145]/[254+399xx253]`

`=[253xx399+399-145]/[254+399xx253]`

`=[254+253xx399]/[254+253xx399]=1`

`b)[5932+6001xx5931]/[5932xx6001-69]`

`=[5932+6001xx5931]/[(5931+1)xx6001-69]`

`=[5932+6001xx5931]/[5931xx6001+6001-69]`

`=[5932+6001xx5931]/[5932+6001xx5931]=1`