\(\left(x+\dfrac{1}{2}\right).\left(x-\dfrac{3}{4}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\x-\dfrac{3}{4}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\)

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

a) \(\left(x-\dfrac{1}{2}\right)^2=0\Rightarrow\left(x-\dfrac{1}{2}\right)^2=0^2\)

\(\Rightarrow x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)

Vậy \(x=\dfrac{1}{2}\)

b) \(\left(x-2\right)^2=1\Rightarrow\left(x-2\right)^2=1^2\)

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

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

c) \(\left(2x-1\right)^3=-8\Rightarrow\left(2x-1\right)^3=-2^3\)

\(\Rightarrow2x-1=-3\Rightarrow2x=-3+1\)

\(\Rightarrow2x=2\Rightarrow x=1\)

Vậy \(x=1\)

d) \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\Rightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{4}\right)^2\)

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

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)

a) Ta có : \(x\in B\left(11\right)=\left\{0;11;22;33;44;55;66;77;...\right\}\)

Mà \(30< x< 70\Rightarrow x\in\left\{33;44;55;66\right\}\)

Vậy \(x\in\left\{33;44;55;66\right\}\)

b) Ta có : \(x\inƯ\left(20\right)=\left\{1;2;4;5;10;20\right\}\)

Mà \(x>4\Rightarrow x\in\left\{5;10;20\right\}\)

Vậy \(x\in\left\{5;10;20\right\}\)

Ta có : \(M=1^2+2^2+3^2+.....+99^2+100^2\)

\(\Rightarrow M=1\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+...99\left(100-1\right)+100.\left(101-1\right)\)

\(\Rightarrow M=1.2-1+2.3-2+3.4-3+...+99.100-99+100.101-100\)

\(\Rightarrow M=\left(1.2+2.3+3.4+...+99.100+100.101\right)-\left(1+2+3+...+99+100\right)\)

Đặt \(A=1.2+2.3+3.4+....+99.100+100.101\)

\(\Rightarrow3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)+100.101.\left(102-99\right)\)

\(\Rightarrow3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100+100.101.102-99.100.101\)

\(\Rightarrow3A=100.101.102\)

\(\Rightarrow A=\dfrac{100.101.102}{3}=343400\) (1)

Đặt \(B=1+2+3+....+99+100\)

\(\Rightarrow B=\left(100+1\right).100:2=5050\) (2)

Từ (1) và (2), suy ra:

\(M=A-B=343400-5050=338350\)

Vậy \(M=338350\)

~ Học tốt ~

Sửa đề: Chứng minh: \(4\left(a-b\right)\left(b-c\right)=4\left(b-c\right)^2\)

Đặt \(\dfrac{a}{2017}=\dfrac{b}{2018}=\dfrac{b}{2019}=k\)

\(\Rightarrow\left\{{}\begin{matrix}a=2017k\\b=2018k\\c=2019k\end{matrix}\right.\)

VT: \(4\left(a-b\right)\left(b-c\right)=4\left(2017k-2018k\right)\left(2018k-2019k\right)\)

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

VP: \(4\left(b-c\right)^2=4\left(2018k-2019k\right)^2=4k^2\) (2)

Từ (1) và (2), suy ra:

\(4\left(a-b\right)\left(b-c\right)=4\left(b-c\right)^2\)\(\Rightarrow\) (đpcm)

~ Học tốt ~

a) Trong 3 điểm O, B, D điểm B nằm giữa hai điểm còn lại. Vì OB < OD (3cm < 7cm).

b) Vì B nằm giữa hai điểm O và D, nên ta có :

OB + BD = OD

\(\Rightarrow\) BD = OD - OB

Hay BD = 7cm - 3cm = 4cm

Vậy BD dài 4cm.

c) Ta có E là trung điểm của BD, nên :

\(BE=ED=\dfrac{BD}{2}\)

Hay \(BE=ED=\dfrac{4cm}{2}=2cm\)

Vậy ED dài 2cm.

~ học tốt ~

Ta có : \(AD=\dfrac{2}{3}AB\Rightarrow\Delta ADC=\dfrac{2}{3}\Delta ABC\)

Hay \(\Delta ADC=900.\dfrac{2}{3}=600\left(cm^2\right)\)

Ta có : \(AE=\dfrac{2}{3}AC\Rightarrow\Delta ADC=\dfrac{2}{3}\Delta ADC\)

Hay \(\Delta ADE=600.\dfrac{2}{3}=400\left(cm^2\right)\)

Vậy diện tích tam giác ADE là: 400 \(cm^2\)

~ Học tốt ~

\(1\dfrac{1}{3};1\dfrac{1}{8};1\dfrac{1}{15};1\dfrac{1}{15};1\dfrac{1}{24};1\dfrac{1}{35};...\)

\(\Leftrightarrow\dfrac{4}{3};\dfrac{9}{8};\dfrac{16}{15};\dfrac{25}{24};\dfrac{36}{35};...\)

\(\Leftrightarrow\dfrac{2^2}{1.3};\dfrac{3^2}{2.4};\dfrac{4^2}{3.5};\dfrac{5^2}{4.6};\dfrac{6^2}{5.7};...\)

\(\Rightarrow\) Tích 100 số hạng đầu tiên của dãy là:

\(\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.\dfrac{5^2}{4.6}.\dfrac{6^2}{5.7}......\dfrac{101^2}{100.102}\)

\(=\dfrac{2.3.4.5.6....101}{1.2.3.4.5.....100}.\dfrac{2.3.4.5.6....101}{3.4.5.6.7....102}\)

\(=101.\dfrac{2}{102}\)\(=\dfrac{101}{51}\)

Vậy..............................

~ học tốt ~

Ta có : \(\dfrac{2017+2018}{2018+2019}=\dfrac{2017}{2018+2019}+\dfrac{2018}{2018+2019}\)

Rõ ràng ta thấy : \(\dfrac{2017}{2018}>\dfrac{2017}{2018+2019}\) (1)

\(\dfrac{2018}{2019}>\dfrac{2018}{2018+2019}\) (2)

Từ (1) và (2), suy ra :

\(\dfrac{2017}{2018}+\dfrac{2018}{2019}>\dfrac{2017+2018}{2018+2019}\)

Vậy ......................

~ Học tốt ~