\(=>x\left(\dfrac{1}{2}+\dfrac{2}{3}\right)=\dfrac{10}{3}+1=\dfrac{13}{3}\)

\(=>x=\dfrac{13}{3}:\left(\dfrac{1}{2}+\dfrac{2}{3}\right)=\dfrac{13}{3}:\dfrac{7}{6}=\dfrac{26}{7}\)

5/6x - 1 = 10/3

5/6x = 10/3 + 1 = 13/3

X = 13/3 : 5/6 = 26/5

\(\dfrac{1}{2}\times x+\dfrac{2}{3}\times x-1=3\dfrac{1}{3}\)

\(\Rightarrow x\left(\dfrac{1}{2}+\dfrac{2}{3}\right)=\dfrac{10}{3}+1\)

\(\Rightarrow\dfrac{7}{6}x=\dfrac{13}{3}\)

\(\Rightarrow x=\dfrac{13}{3}:\dfrac{7}{6}\)

\(\Rightarrow x=\dfrac{26}{7}\)

`1/2x^2(-2x^2y^2z).(-1/3)x^2y^3`

`=1/2 .(-2).(-1/3)x^{2+2+2}.y^{2+3}.z`

`=1/3x^6y^5z`

Ta có: \(\dfrac{1}{2}x^2\cdot\left(-2x^2y^2z\right)\cdot\left(-\dfrac{1}{3}\right)x^2y^3\)

\(=\left(\dfrac{1}{2}\cdot2\cdot\dfrac{1}{3}\right)\cdot\left(x^2\cdot x^2\cdot x^2\right)\cdot\left(y^2\cdot y^3\right)\cdot z\)

\(=\dfrac{1}{3}x^6y^5z\)

\(\dfrac{5}{7}\times x+\dfrac{2}{5}=\dfrac{2}{3}\\ \dfrac{5}{7}\times x=\dfrac{2}{3}-\dfrac{2}{5}=\dfrac{10}{15}-\dfrac{6}{15}=\dfrac{4}{15}\\ x=\dfrac{4}{15}:\dfrac{5}{7}=\dfrac{4\times7}{15\times5}=\dfrac{28}{75}\)

Vậy `x=28/75`

`HaNa☘D`

\(\dfrac{5}{7}\times x+\dfrac{2}{5}=\dfrac{2}{3}\)

\(\dfrac{5}{7}\times x=\dfrac{2}{3}-\dfrac{2}{5}\)

\(\dfrac{5}{7}\times x=\dfrac{10}{15}-\dfrac{6}{15}\)

\(\dfrac{5}{7}\times x=\dfrac{4}{15}\)

\(x=\dfrac{4}{15}:\dfrac{5}{7}\)

\(x=\dfrac{4}{15}\times\dfrac{7}{5}\)

\(x=\dfrac{28}{75}\)

#Toru

=>x*5/7=2/3-2/5=4/15

=>x=4/15:5/7=4/15*7/5=28/75

\(\dfrac{-3}{2}\times\dfrac{-15}{64}+\left(0,8-2\dfrac{4}{5}\right):3\dfrac{2}{3}\\ =\dfrac{45}{128}+\left(-2\right)\times\dfrac{3}{11}\\ =\dfrac{45}{128}-\dfrac{6}{11}\\ =-\dfrac{273}{1408}\)

\(x^3-9x^2=x^3-9x^2+2x-3\)

⇔\(x^3-x^3-9x^2+9x^2-2x=-3\)

⇔\(-2x=-3\)

⇔\(x=\dfrac{3}{2}\)

`x^3 -9x^2 =x^3 -9x^2 +2x-3`

`=> x^3 -9x^2-x^3 +9x^2 -2x+3=0`

`=> (x^3-x^3) +(-9x^2+9x^2)-2x+3=0`

`=> -2x-3=0`

`=>-2x=-3`

`=>x=3/2`

ban co the bao minh cach lam ko

Ta có: \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)

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

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

\(\Rightarrow x-1=\left\{-1;1\right\}\)

Nếu x - 1 = - 1 thì x = 0

Nếu x - 1 = 1 thì x = 2

Vậy x mang 2 giá trị là: x =2 hoặc x = 0  

Bạn Duong làm thiếu nghiệm.

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

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

\(\Leftrightarrow\left(x-1\right)^{x+2}\left[1-\left(x-1\right)^4\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^{x+2}=0\\\left(x-1\right)^4=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x-1=1;x-1=-1\end{cases}}\)

 \(\Leftrightarrow x=1;x=2;x=0\)

\(∘backwin\)

\(a ) ( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ... + ( x + 100 ) = 5750\)

\( ( x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 100 ) = 5750 \)

\( 100 x + ( 1 + 100 ) ×100 : 2 = 5750\)

\(100 x + 5050 = 5750\)

\( 100 x = 5750 − 5050\)

\(100 x = 700\)

\(x = 700 : 100\)

\(x = 7\)

\(b,\) \(B=\)\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2021^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2020}+2021\)

\( B < 1 -\)\(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2020}-\dfrac{1}{2021}\)

\(B<1-\)\(\dfrac{1}{2021}\)

\(B<\)\(\dfrac{2020}{2021}\)

\(\dfrac{2020}{2021}< 1\)

\(B<1\)

a) (x+1) +(x+2 ) + ...+(x+100)=5750
= 100x + (1+2+3+...+100) = 5750
=100x + 5050 = 5750
--> 100x = 5750-5050=700
--> x=7

b) Ta thấy: 1/2^2 < 1/2.3
                  1/3^2 < 1/3.4
                        ...
                  1/2021^2 < 1/2021.2022
--> B=1/2^2 + 1/3^2 + 1/4^2 + ...+ 1/2021^2 < 1/2.3 + 1/3.4 + ... +1/2021.2022 (1)
     Ta có: 1/2.3 + 1/3.4 + ... +1/2021.2022
         =1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2021 - 1/2022
         =1/2 - 1/2022 < 1 (2)
Từ (1) và (2) --> B<1 (đpcm)
                                                                      < 
                                                                     

\(3\frac{3}{11}x27\frac{27}{46}x1\frac{6}{17}x2\frac{4}{9}=\frac{34x1269x23x22}{11x46x17x9}=\frac{2x17x141x9x23x2x11}{11x23x17x9}=282\)

282 la dung chac luon

282 đó bạn nhé