b) Ta có : \(\left(x-\frac{1}{3}\right)^2-\frac{1}{4}=0\)

\(\Rightarrow\left(x-\frac{1}{3}\right)^2=\frac{1}{4}\)

\(\Rightarrow\orbr{\begin{cases}\left(x-\frac{1}{3}\right)^2=\left(\frac{1}{2}\right)^2\\\left(x-\frac{1}{3}\right)^2=\left(-\frac{1}{2}\right)^2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{1}{2}\\x-\frac{1}{3}=-\frac{1}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{6}\\x=-\frac{1}{6}\end{cases}}\)

b) \(\left(x-\frac{1}{3}\right)^2-\frac{1}{4}=0\)

\(\Leftrightarrow\left(x-\frac{1}{3}\right)^2=\frac{1}{4}\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{1}{4}\\x-\frac{1}{3}=-\frac{1}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{12}\\x=\frac{1}{12}\end{cases}}\)

d) \(\frac{x+5}{2}=\frac{8}{x+5}\)

\(\Rightarrow\left(x+5\right)^2=16\)

\(\Rightarrow\orbr{\begin{cases}x+5=16\\x+5=-16\end{cases}\Rightarrow\orbr{\begin{cases}x=11\\x=-21\end{cases}}}\)

a, \(\dfrac{x-2}{5}=\dfrac{x}{3}\)

\(\Leftrightarrow3\left(x-2\right)=5x\)

\(\Leftrightarrow3x-6=5x\)

\(\Leftrightarrow5x-3x=6\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=3\)

b, \(\dfrac{x+23}{x+40}=\dfrac{3}{4}\)

\(\Leftrightarrow4\left(x+23\right)=3\left(x+40\right)\)

\(\Leftrightarrow4x+92=2x+80\)

\(\Leftrightarrow4x-2x=80-92\)

\(\Leftrightarrow2x=-12\)

\(\Leftrightarrow x=-6\)

c, \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...........+\dfrac{1}{2^{2017}}\)

\(\Leftrightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...........+\dfrac{1}{2^{2016}}\)

\(\Leftrightarrow2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+........+\dfrac{1}{2^{2016}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+........+\dfrac{1}{2^{2017}}\right)\)

\(\Leftrightarrow A=1-\dfrac{1}{2^{2017}}\)

d, \(B=1+2+2^2+........+2^{2017}\)

\(\Leftrightarrow2B=2+2^2+2^3+......+2^{2018}\)

\(\Leftrightarrow2B-B=\left(2+2^2+.....+2^{2018}\right)-\left(1+2+....+2^{2017}\right)\)

\(\Leftrightarrow B=2^{2018}-1\)

\(\dfrac{x-2}{5}=\dfrac{x}{3}=>3\left(x-2\right)=5x\)

\(< =>3x-6=5x=>x=-3\)

\(\dfrac{x+23}{x+40}=\dfrac{3}{4}=>4\left(x+23\right)=3\left(x+40\right)\)

\(4x+92=3x+120=>x=28\)

a)

\(x+\left(x-1\right)+\left(x-2\right)+...+\left(x-50\right)=255\\ x+x-1+x-2+...+x-50=255\\ \left(x+x+x+...+x\right)-\left(1+2+3+...+50\right)\\ 51x-1275=255\\ 51x=1530\\ x=30\)

e)

\(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+30\right)=1240\\ x+x+1+x+2+...+x+30=1240\\ \left(x+x+x+...+x\right)+\left(1+2+3+...+30\right)=1240\\ 31x+465=1240\\ 31x=775\\ x=25\)

f)

\(\left(x-1\right)+\left(x-2\right)+...+\left(x-19\right)+\left(x-20\right)=-610\\ x-1+x-2+...+x-19+x-20=-610\\ \left(x+x+x+...+x\right)-\left(1+2+3+...+20\right)=-610\\ 20x-210=-610\\ 20x=-400\\ x=-20\)

\(3^{x+2}+3^x=90\Leftrightarrow3^x.3^2+3^x=90\Leftrightarrow3^x\left(3^2+1\right)=90\Leftrightarrow3^x.10=90\Leftrightarrow3^x=9\Leftrightarrow3^x=3^2\Leftrightarrow x=2\)

Vậy ...

\(\left(x+\dfrac{1}{2}\right)^2-\dfrac{1}{3}=\dfrac{1}{9}\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{4}{9}\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{2}{3}\right)^2\Leftrightarrow x+\dfrac{1}{2}=\dfrac{2}{3}\Leftrightarrow x=\dfrac{1}{6}\)

Vậy ...

1.

a) \(3^{x+2}+3^x=90\)

\(\Leftrightarrow3^x\left(3^2+1\right)=90\)

\(\Leftrightarrow3^x.10=90\)

\(\Leftrightarrow3^x=9=3^2\)

\(\Leftrightarrow x=2\)

vậy...

b) \(\left(x+\dfrac{1}{2}\right)^2-\dfrac{1}{3}=\dfrac{1}{9}\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{4}{9}\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\pm\dfrac{2}{3}\right)^2\)

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

Vậy...

tik mik nha !!!

a: =>5x=3x-6

=>2x=-6

hay x=-3

b: \(\Leftrightarrow\left(x-3\right)^2=4\cdot5^2=100\)

=>x-3=10 hoặc x-3=-10

=>x=13 hoặc x=-7

c: \(\left|x^3+1\right|+2\ge2\forall x\)

Dấu '=' xảy ra khi x=-1

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

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

\(\Leftrightarrow3x-6=5x\)

\(\Leftrightarrow5x-3x=-6\)

\(\Leftrightarrow2x=-6\)

\(\Leftrightarrow x=-3\)

Vậy .....

b, \(B=1+2+2^2+..........+2^{2017}\)

\(\Leftrightarrow2B=2+2^2+.......+2^{2018}\)

\(\Leftrightarrow2B-B=\left(2+2^2+......+2^{2018}\right)-\left(1+2+......+2^{2017}\right)\)

\(\Leftrightarrow B=2^{2018}-1\)

c, \(\dfrac{x+23}{x+40}=\dfrac{3}{4}\)

\(\Leftrightarrow4\left(x+23\right)=3\left(x+40\right)\)

\(\Leftrightarrow4x+92=3x+120\)

\(\Leftrightarrow4x-3x=120-92\)

\(\Leftrightarrow x=28\)

\(\left|2x\right|+2x=0\)

\(\Rightarrow\left|2x\right|=-2x\)

\(\Rightarrow2x\le0\)

\(\Rightarrow x\le0\)

Vậy \(x\le0\)

\(\left(x-1\right).\left(x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}}\)

Vậy \(\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)

\(\left|x-3\right|+x-3=0\)

\(\left|x-3\right|=-x+3\)

\(\left|x-3\right|=-\left(x-3\right)\)

\(\Rightarrow x-3\le0\)

\(\Rightarrow x\le3\)

Vậy \(x\le3\)

\(\left(x+1\right)^3=\left(x+1\right)^5\)

\(\left(x+1\right)^5-\left(x+1\right)^3=0\)

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

\(\orbr{\begin{cases}\left(x+1\right)^3=0\\\left(x+1\right)^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=0\end{cases}}}\)hoặc \(x=-2\)

Vậy \(x\in\left\{-1;0;-2\right\}\)

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

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

\(\Rightarrow x-2=2^3\)

\(x=8+2\)

\(x=10\)

Vậy \(x=10\)

Câu 6 tương tự câu 4

Tham khảo nhé~

P/S: nên chia nhỏ đăng thành nhiều bài khác nhau

b) \(\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}=0\)

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

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

Vây: \(x=0;1\)

_Chúc bạn học tốt_

\(\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{59.61}\)

= \(\dfrac{2}{2}.\left(\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{59.61}\right)\)

= \(\dfrac{3}{2}.\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\right)\)

= \(\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\)

= \(\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{61}\right)\)

=\(\dfrac{3}{2}.\dfrac{56}{305}\)

= \(\dfrac{78}{305}\)

\(\left(x^2-4\right)\left(6-2x\right)=0\) ⇔ \(x^2-4=0\) hoặc \(6-2x=0\)

*Nếu \(x^2-4=0\)

⇒ x² = 4

⇒ x ∈ {2 ; -2}

*Nếu \(6-2x=0\)

⇒2x = 6

⇒ x = 6 : 2 = 3

Vậy x ∈ { -2 ; 2 ; 3 }

1.Tính hợp lý:

a. 1152 - (374 + 1152) + (374 - 65) = 1152 - 374 - 1152 + 374 - 65 = ( 1152 - 1152 ) + ( -65) + ( 374 - 374 ) = 0 + ( - 65) + 0 = -65

Bài 1 : Tính hợp lý : c. \(\dfrac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\) = \(\dfrac{11.3^{29}-3^{30}}{2^2.3^{28}}\) = \(\dfrac{3^{29}.\left(11-3\right)}{2^2.3^{28}}\) = \(\dfrac{3^{29}.2^3}{2^2.3^{28}}\) = 6