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1)\(3,5\cdot x+1,54\cdot3,5=15\)

    \(3,5\left(x+1,54\right)=15\)

                 \(x+1,54=\frac{15}{3,5}\)

                 \(x+1,54=\frac{30}{7}\)

                                \(x=\frac{30}{7}-1,54\)

                                 \(x=\frac{30}{7}-\frac{77}{50}\)

                               \(x=\frac{961}{350}\)

\(20+\frac{12}{x-2019}=32\)

             \(\frac{12}{x-2019}=32-20\)

              \(\frac{12}{x-2019}=12\)

               \(x-2019=\frac{12}{12}\)

              \(x-2019=1\)

                               \(x=1+2019\)

2)

\(\frac{3}{7}\cdot\frac{4}{13}+\frac{3}{7}\cdot\frac{9}{13}+5\frac{4}{7}\)

\(=\frac{3}{7}\cdot\frac{4}{13}+\frac{3}{7}\cdot\frac{9}{13}+13\cdot\frac{3}{7}\)

\(=\frac{3}{7}\left(\frac{4}{13}+\frac{9}{13}+13\right)\)

\(=\frac{3}{7}\cdot14\)

\(=6\)

17 tháng 7 2020

3.5*(x-1.54)=15

x+1.54=15;3.5

x+1.54=4.28

x=4.28-1.54

x=2.74

chưa suy nghĩ

26 tháng 12 2023

a) \(\left(x-3\right)^2+\left(4-x\right)\left(x+4\right)=10\)

\(\Leftrightarrow\left(x^2-2\cdot x\cdot3+3^2\right)+\left(4-x\right)\left(4+x\right)=10\)

\(\Leftrightarrow x^2-6x+9+\left(4^2-x^2\right)-10=0\)

\(\Leftrightarrow x^2-6x-1+16-x^2=0\)

\(\Leftrightarrow-6x+15=0\)

\(\Leftrightarrow6x=15\)

\(\Leftrightarrow x=\dfrac{5}{2}\)

b) \(x^2-2x=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

c) \(\left(x^2-9\right)^2-\left(x-3\right)^2=0\)

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

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

\(\Leftrightarrow\left(x-3\right)^2\left[\left(x+3\right)^2-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-3\right)^2=0\\\left(x+3\right)^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-3=0\\\left(x+3\right)^2=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x+3=1\\x+3=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\\x=-4\end{matrix}\right.\)

27 tháng 12 2022

\(\dfrac{4}{5}-\left(-\dfrac{2}{7}\right)-\dfrac{7}{10}\\ =\dfrac{4}{5}+\dfrac{2}{7}-\dfrac{7}{10}\\ =\dfrac{56}{70}+\dfrac{20}{70}-\dfrac{49}{70}\\ =\dfrac{27}{70}\\ 3,5-\left(-\dfrac{2}{7}\right)\\ =\dfrac{7}{2}+\dfrac{2}{7}\\ =\dfrac{49}{14}+\dfrac{4}{14}\\ =\dfrac{53}{14}\\ \left(-3\right).\left(-\dfrac{7}{12}\right)\\ =\dfrac{21}{12}\\ =\dfrac{7}{4}\)

4 tháng 12 2021

D

4 tháng 12 2021

d

16 tháng 6 2023

a) 6 x 3  =18               7 x 5 =35                     9 x 4=36              8 x 10=80

b) 8 x 7 =56                3 x 9 =27                    5 x 6=30              4 x 8=32

16 tháng 6 2023

a) 6 x 3=18                 7 x 5=35                     9 x 4=36              8 x 10=80

b) 8 x 7=56                 3 x 9=27                     5 x 6=30              4 x 8=32

14 tháng 5 2018

Câu đầu:

12/5 - X x 3/4 =5/4 

X x 3/4 =12/5 - 5/4 ra X x 3/4 = 23/20 , X = 23/20 : 3/4 ,X=23/15

Câu hai:

X : 3/5 + 3/8 =15/8

X : 3/5 =15/8 - 3/8 ra X : 3/5 = 15/8 , X= 15/8 x 3/5,X= 9/8

27 tháng 2 2022

hahahaaaaaaaaaaaaaaaaaaaaaa

23 tháng 7 2023

a) \(\left(x+1\right)^3-\left(x-1\right)^3-6\cdot\left(x-1\right)^2=10\)

\(\Rightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\cdot\left(x^2-2x+1\right)=10\)

\(\Rightarrow6x^2+2-6x^2+12x-6=10\)

\(\Rightarrow12x-4=10\)

\(\Rightarrow12x=14\)

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

b) \(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=42\)

\(\Rightarrow x\left(x^2-25\right)-\left(x^3+8\right)=42\)

\(\Rightarrow x^3-25x-x^3-8=42\)

\(\Rightarrow-25x-8=42\)

\(\Rightarrow-25x=50\)

\(\Rightarrow x=\dfrac{50}{-25}=-2\)

c) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)

\(\Rightarrow x^3-6x^2+12x-8-\left(x^3-27\right)+6\left(x^2+2x+1\right)=49\)

\(\Rightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)

\(\Rightarrow24x+25=49\)

\(\Rightarrow24x=24\)

\(\Rightarrow x=\dfrac{24}{24}=1\)

24 tháng 10 2021

a) \(x\in\left\{15;30;45\right\}\)

b) \(x\in\left\{12;24;36\right\}\)

c) \(x\in\left\{60;120\right\}\)

30 tháng 6 2021

a) ⇔ |2x+3| = 8
⇒ \(\left[{}\begin{matrix}2x+3=8\\2x+3=-8\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}2x=5\\2x=-11\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{11}{2}\end{matrix}\right.\)

Vậy...

30 tháng 6 2021

b) ĐKXĐ: \(x\ge0\)

\(\Leftrightarrow3\sqrt{x}-7\sqrt{x}+6\sqrt{x}=8\)

\(\Leftrightarrow2\sqrt{x}=8\)

\(\Leftrightarrow\sqrt{x}=4\)

\(\Leftrightarrow x=16\) (Vì \(x\ge0\) )

Vậy x = 16

c) ĐKXĐ: \(x\ge1\)

\(\Leftrightarrow\sqrt{9\left(x-1\right)}=12\)

\(\Leftrightarrow3\sqrt{x-1}=12\)

\(\Leftrightarrow\sqrt{x-1}=4\)

\(\Leftrightarrow x-1=16\)

\(\Leftrightarrow x=17\)(TM)

Vậy x = 17

a) Ta có: \(9\cdot5^x=6\cdot5^6+3\cdot5^6\)

\(\Leftrightarrow9\cdot5^x=9\cdot5^6\)

\(\Leftrightarrow5^x=5^6\)

hay x=6

b) Ta có: \(2^{2x+1}+4^{x+3}=264\)

\(\Leftrightarrow4^x\cdot2+4^x\cdot64=264\)

\(\Leftrightarrow4^x=4\)

hay x=1