Số cần tìm là \(26:\dfrac{2}{3}=26\cdot\dfrac{3}{2}=39\)

giúp tuiiiiiiiiiiii

p: \(30x-3x=5\cdot54\)

=>\(27x=270\)

=>\(x=\dfrac{270}{27}=10\)

q: 3(x-2)+2(x+5)=29

=>3x-6+2x+10=29

=>5x+4=29

=>5x=25

=>\(x=\dfrac{25}{5}=5\)

t: (27-3x)(x-5)=0

=>3(9-x)(x-5)=0

=>(9-x)(x-5)=0

=>\(\left[{}\begin{matrix}9-x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=5\end{matrix}\right.\)

v: x+(x+1)+...+(x+30)=1240

=>31x+(1+2+...+30)=1240

=>\(31x+30\cdot\dfrac{31}{2}=1240\)

=>\(31\left(x+15\right)=31\cdot40\)

=>x+15=40

=>x=40-15=25

s: (x+2)+(4x+4)+(7x+6)+...+(25x+18)+(28x+20)=1560

=>(x+4x+7x+...+25x+28x)+(2+4+6+...+20)=1560

=>\(x\left(1+4+...+28\right)+2\left(1+2+3+...+10\right)=1560\)

=>\(x\left[\left(\dfrac{28-1}{3}+1\right)\cdot\dfrac{\left(28+1\right)}{2}\right]+2\cdot\dfrac{10\cdot11}{2}=1560\)

=>\(x\left[10\cdot\dfrac{29}{2}\right]+10\cdot11=1560\)

=>\(145x=1560-110=1450\)

=>x=10

a: \(2^x+2^{x+4}=544\)

=>\(2^x+2^x\cdot16=544\)

=>\(17\cdot2^x=544\)

=>\(2^x=32=2^5\)

=>x=5

b: \(4^{2x+1}+4^{2x}=80\)

=>\(4^{2x}\cdot4+4^{2x}=80\)

=>\(4^{2x}\cdot5=80\)

=>\(4^{2x}=16=4^2\)

=>2x=2

=>x=1

c: \(3^{2x+2}+3^{2x+1}=108\)

=>\(3^{2x}\cdot9+3^{2x}\cdot3=108\)

=>\(12\cdot3^{2x}=108\)

=>\(3^{2x}=9=3^2\)

=>2x=2

=>x=1

d: \(7^{x+3}-7^{x+1}=16464\)

=>\(7^x\cdot343-7^x\cdot7=16464\)

=>\(7^x\cdot336=16464\)

=>\(7^x=49=7^2\)

=>x=2

BM+MN+NC=BC
=>\(BC=MN+\dfrac{2}{3}MN+\dfrac{1}{2}MN=\dfrac{13}{6}MN\)

=>\(\dfrac{BM}{BC}=\dfrac{2}{3}:\dfrac{13}{6}=\dfrac{2}{3}\cdot\dfrac{6}{13}=\dfrac{12}{39}=\dfrac{4}{13}\)

=>BC=3,25BM

\(S_{ABM}=\dfrac{1}{2}\cdot MH\cdot AB=\dfrac{1}{2}\cdot12\cdot25=150\left(cm^2\right)\)

BC=3,25BM nên \(S_{ABC}=3,25\cdot S_{ABM}=3,25\cdot150=487,5\left(cm^2\right)\)

Gọi số cần tìm là ab (a khác 0; a, b là chữ số)

Ta có:

ab = b x b +a

=> 10 x a +b = b x b +a

=> 9 x a = b x [b-1]

Vì b , b-1 là hai số tự nhiên liên tiếp nên a và 9 là hai số tự nhiên liên tiếp =>a = 8

=> b x [b-1] = 72 => b= 9

Vậy số cần tìm là 89

\(\dfrac{x+5}{2005}+\dfrac{x+6}{2004}+\dfrac{x+7}{2003}=-3\\ \Rightarrow\dfrac{x+5}{2005}+1+\dfrac{x+6}{2004}+1+\dfrac{x+7}{2003}+1=0\\ \Rightarrow\dfrac{x+2010}{2005}+\dfrac{x+2010}{2004}+\dfrac{x+2010}{2003}=0\\ \Rightarrow\left(x+2010\right)\left(\dfrac{1}{2005}+\dfrac{1}{2004}+\dfrac{1}{2003}\right)=0\\ \Rightarrow x+2010=0\\ \Rightarrow x=-2010\)

b) \(A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}+\dfrac{3}{97.100}\\ =1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{99}{100}\)

\(\overline{ab}=b^2+a\)

\(10a+b=b^2+a\)

\(9a=b^2-b\)

\(a=\dfrac{b\left(b-1\right)}{9}\)

\(b=9,a=8\)

Gọi số cần tìm là \(\overline{ab}\)

Ta có:\(\overline{ab}=b^2+a\)

Mà:\(\overline{ab}=10a+b\)

\(\Rightarrow b^2+a=10a+b\)

\(\Rightarrow b^2-b=10a-a\)

\(\Rightarrow b\left(b-1\right)=9a\)

Vì \(b\left(b-1\right)\) là \(2\) số tự nhiên liên tiếp nên \(9a\) là \(2\) số tự nhiên liên tiếp

\(\Rightarrow\left[{}\begin{matrix}a=8\\a=10\end{matrix}\right.\)

Mà \(a< 10\)

\(\Rightarrow a=8\)

\(\Rightarrow b\left(b-1\right)=9.8\)

\(\Rightarrow b=9\)

Vậy số cần tìm là \(89\)

`#3107.101107`

`8)`

\(15\cdot24-14\cdot5\left(145\right)\div5-27\\ =15\cdot24-14\cdot145-27\\ =360-2030-27\\ =-1697\)

Vẽ hình vuông thì cần gì compa em ơi?

\(\left(\dfrac{2}{3}x-\dfrac{4}{9}\right)\left(\dfrac{1}{2}-\dfrac{3}{7}:x\right)=0\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x-\dfrac{4}{9}=0\\\dfrac{1}{2}-\dfrac{3}{7}:x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=\dfrac{4}{9}\\\dfrac{3}{7}:x=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{6}{7}\end{matrix}\right.\)