Lời giải:

Đặt $A=3+3^2+3^3+...+3^{100}$

$3A=3^2+3^3+3^4+...+3^{101}$

$\Rightarrow 2A=3A-A=(3^2+3^3+3^4+...+3^{101})-(3+3^2+3^3+...+3^{100})$

$\Rightarrow 2A=3^{101}-3$

Có: $2A=3^x-5$

$\Rightarrow 3^{101}-3=3^x-5$

$\Rightarrow 3^{101}=3^x-2$

Giá trị $x$ khi đó tìm được sẽ không phù hợp với lớp 6. Bạn xem lại đề.

$2A=3^x-5$

$\Rightarrow 3^{101}-

3 + x - 5 = 6x -4

3 - 5 + 4 = 6x -x 

2             = 5x 

x = 2 / 5 

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

\(3+x-5=6x-4\)

\(x-6x=-4+5-3\)

\(-5x=-2\)

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

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

3 + ( x - 5 ) = 2 . 3x - 2. 2

3 + x + ( -5 ) = 6x - 4

3 + ( -5 ) + x = 6x - 4

-2 + x = 6x - 4

- 6x + x = 2 - 4

- 5 x = -2

x = -2 : ( -5 ) 

x = 0,4

\(a,\frac{2}{7}:\frac{4}{5}=\frac{2}{7}\times\frac{5}{4}=\frac{10}{28}=\frac{5}{14}\)

\(b,\frac{3}{8}:\frac{9}{4}=\frac{3}{8}\times\frac{4}{9}=\frac{12}{72}=\frac{1}{6}\)

\(c,\frac{8}{21}:\frac{4}{7}=\frac{8}{21}\times\frac{7}{4}=\frac{56}{84}=\frac{2}{3}\)

\(d,\frac{5}{8}:\frac{15}{8}=\frac{5}{8}\times\frac{8}{15}=\frac{40}{120}=\frac{1}{3}\)

A,\(\frac{2}{7}:\frac{4}{5}=\frac{2}{7}\times\frac{5}{4}=\frac{10}{28}=\frac{5}{14}\)

B,\(\frac{3}{8}:\frac{9}{4}=\frac{3}{8}\times\frac{4}{9}=\frac{12}{72}=\frac{1}{6}\)

C.\(\frac{8}{21}:\frac{4}{7}=\frac{8}{21}\times\frac{7}{4}=\frac{56}{84}=\frac{2}{3}\)

D.\(\frac{5}{8}:\frac{15}{8}=\frac{5}{8}\times\frac{8}{15}=\frac{40}{120}=\frac{1}{3}\)

Đúng 100%

a, \(\frac{2}{7}:\frac{4}{5}=\)\(\frac{2}{7}x\)\(\frac{5}{4}\)\(\frac{10}{28}=\frac{5}{14}\)

b,\(\frac{3}{8}:\frac{9}{4}\)\(=\frac{3}{8}x\)\(\frac{4}{9}\)\(=\frac{12}{72}=\frac{1}{6}\)

c,\(\frac{8}{21}:\frac{4}{7}\)\(=\frac{8}{21}x\)\(\frac{7}{4}\)\(=\frac{56}{84}=\frac{2}{3}\)

d,\(\frac{5}{8}:\frac{15}{8}\)\(\frac{5}{8}x\)\(\frac{8}{15}=\frac{40}{120}=\frac{1}{3}\)

tk giup mk nha

\(\dfrac{x+1}{199}+\dfrac{x+2}{198}+\dfrac{x+3}{197}+\dfrac{x+4}{196}+\dfrac{x+220}{5}=0\)

\(\Leftrightarrow\left(\dfrac{x+1}{199}+1\right)+\left(\dfrac{x+2}{198}+1\right)+\left(\dfrac{x+3}{197}+1\right)+\left(\dfrac{x+4}{196}+1\right)+\dfrac{x+200}{5}+\dfrac{20}{5}-4=0\)

\(\Leftrightarrow\dfrac{x+200}{199}+\dfrac{x+200}{198}+\dfrac{x+200}{197}+\dfrac{x+200}{196}+\dfrac{x+200}{5}=0\)

\(\Leftrightarrow\left(x+200\right)\left(\dfrac{1}{199}+\dfrac{1}{198}+\dfrac{1}{197}+\dfrac{1}{196}+\dfrac{1}{5}\right)=0\)

\(\Leftrightarrow x=-200\)( do \(\dfrac{1}{199}+\dfrac{1}{198}+\dfrac{1}{197}+\dfrac{1}{196}+\dfrac{1}{5}>0\))

\(\dfrac{x+1}{199}+\dfrac{x+2}{198}+\dfrac{x+3}{197}+\dfrac{x+4}{196}+\dfrac{x+220}{5}=0\\ \Leftrightarrow\left(\dfrac{x+1}{199}+1\right)+\left(\dfrac{x+2}{198}+1\right)+\left(\dfrac{x+3}{197}+1\right)+\left(\dfrac{x+4}{196}+1\right)+\left(\dfrac{x+220}{5}-4\right)=0\\ \Leftrightarrow\dfrac{x+200}{199}+\dfrac{x+200}{198}+\dfrac{x+200}{197}+\dfrac{x+200}{196}+\dfrac{x+200}{5}=0\\ \Leftrightarrow\left(x+200\right)\left(\dfrac{1}{199}+\dfrac{1}{198}+\dfrac{1}{197}+\dfrac{1}{196}+\dfrac{1}{5}\right)=0\\ \Leftrightarrow x=-200\)

1) \(2x\cdot\left(x-3\right)-5=3x\left(2x-5\right)-4x^2+40\)

\(\Leftrightarrow2x^2-6x-5=6x^2-15x-4x^2+40\)

\(\Leftrightarrow2x^2-6x-5=2x^2-15x+40\)

\(\Leftrightarrow2x^2-6x-5-2x^2+15x-40=0\)

\(\Leftrightarrow9x-45=0\)

<=> x=5

2) x(2x-1)-5(-7)²=2x²-2x+5

<=> 2x²-x-5.49=2x²-2x+5

<=> 2x²-x-245-2x²+2x-5=0

<=> x-250=0

<=> x=250

3) |a-2|=10

\(\Leftrightarrow\orbr{\begin{cases}x-2=10\\x-2=-10\end{cases}\Leftrightarrow\orbr{\begin{cases}x=12\\x=-8\end{cases}}}\)

4) |x|=-5

=> Không tồn tại giá trị của x thỏa mãn vì |x| >=0 với mọi x thuộc Z

Đề là gì vậy ???

\(B=x\left(x-2\right)\left(x+2\right)-\left(x+3\right)\left(x^2-3x+9\right)\)

\(B=x\left(x^2-4\right)-\left(x^3-3x^2+9x+3x^2-9x+27\right)\)

\(B=x^3-4x-\left(x^3+27\right)\)

\(B=-4x-27\)

đề là rút gọn ạ

a. 3^x=1-x^2

x=0 la nghiem

x>=1;  VT>=3 VP<=0 vo nghiem 

b. (de bai thieu n khac 0 vi neu n=0 dung voi moi x)

3x-14=1=> x=5

c.(5^2x5^^x+1)=5^4

5^{^x+1}=5^2=> x=1

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=\dfrac{-3x-4y+5z+3-12-25}{-3\cdot2-4\cdot4+5\cdot6}=\dfrac{16}{8}=2\)

Do đó: x=5; y=5; z=17