\(\left(x+2\right)^2-x^2+4=0\)

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

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

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

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

\(\Leftrightarrow4\left(x+2\right)=0\)

\(\Leftrightarrow x+2=0\)

\(\Leftrightarrow x=-2\)

tìm a,b,c,d thỏa mãn

a²-2a+b²+4b+4c²-4c+6=0

2n+5chia hết cho 2n+1

=>4n+10chia hết cho 4n+2

=>2n+5chia hết cho 2n+1

Ta có: 2n + 5 = (2n - 1) + 6

Do 2n - 1 \(⋮\)2n - 1 => 6 \(⋮\)2n - 1

=> 2n - 1 \(\in\)Ư(6) = {1; 2; 3; 6}

=> 2n \(\in\){2; 3; 4; 7}

Do n \(\in\)N=> n \(\in\){1; 2}

Mình k cho bạn Edogawa Cona rùi nhé.Thanks

tính như bth thôi ráng lên bạn nhá

99/32

\(y x 4/9 - 5/8 = 1/2 : 2/3\)

\(y x 4/9 - 5/8 = 3/4\)

\(y x 4/9 = 3/4 + 5/8\)

\(y x 4/9 = 11/8\)

\(y = 11/8 : 4/9\)

\(y = 99/32\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{3}{5}>\dfrac{2}{5}\\\dfrac{1}{2}x-\dfrac{3}{5}< -\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x>1\\\dfrac{1}{2}x< \dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>2\\x< \dfrac{2}{5}\end{matrix}\right.\)

\(\Rightarrow x+x+...+x+1+2+...+20=2023\)

\(\Rightarrow10x+20.21:2=2023\Rightarrow10x+210=2023\Rightarrow10x=1813\Rightarrow x=\dfrac{1813}{10}\)

1813/10

a) \(\left(\frac{11}{12}+\frac{11}{12.23}+\frac{11}{23.34}+...+\frac{11}{89.100}\right)-x=\frac{2}{3}\)

\(\left(1-\frac{1}{12}+\frac{1}{12}-\frac{1}{23}+\frac{1}{23}-\frac{1}{34}+...+\frac{1}{89}-\frac{1}{100}\right)-x=\frac{2}{3}\)

\(\left(1-\frac{1}{100}\right)-x=\frac{2}{3}\)

\(\frac{99}{100}-x=\frac{2}{3}\)

\(x=\frac{99}{100}-\frac{2}{3}\)

\(x=\frac{97}{300}\)

b) \(\frac{x+1}{9}+\frac{x+3}{7}+\frac{x+5}{5}+\frac{x+7}{3}=a\)

\(\Rightarrow\frac{x+1}{9}+1+\frac{x+3}{7}+1+\frac{x+5}{5}+1+\frac{x+7}{3}+1=a+4\)

\(\frac{x+10}{9}+\frac{x+10}{7}+\frac{x+10}{5}+\frac{x+10}{3}=a+4\)

\(\left(x+10\right).\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)=a+4\)

\(a,\left(\frac{11}{12}+\frac{11}{12\cdot23}+\frac{11}{23\cdot34}+...+\frac{11}{89\cdot100}\right)-x=\frac{2}{3}\)

\(\Rightarrow\left(1-\frac{1}{12}+\frac{1}{12}-\frac{1}{23}+\frac{1}{23}-\frac{1}{34}+...+\frac{1}{89}-\frac{1}{100}\right)-x=\frac{2}{3}\)

\(\Rightarrow1-\frac{1}{100}-x=\frac{2}{3}\)

\(\Rightarrow\frac{99}{100}-x=\frac{2}{3}\)

\(\Rightarrow x=\frac{99}{100}-\frac{2}{3}\)

\(\Rightarrow x=\frac{97}{300}\)

b, k hiểu đề :v

\(x-\frac{1}{4}=\frac{3}{5}\)

\(x=\frac{3}{5}+\frac{1}{4}\)

\(x=\frac{12}{20}+\frac{5}{20}\)

\(x=\frac{17}{20}\)

\(x:\frac{2}{3}=\frac{1}{2}\)

\(x=\frac{1}{2}.\frac{2}{3}\)

\(x=\frac{1}{3}\)

Hok tốt

\(x-\frac{1}{4}=\frac{3}{5}\)

\(x\)        \(=\frac{3}{5}+\frac{1}{4}\)

\(x\)          \(=\frac{12}{20}+\frac{5}{20}\)

\(x\)           \(=\frac{17}{20}\)

\(x:\frac{2}{3}=\frac{1}{2}\)

\(x\)    \(=\frac{1}{2}x\frac{2}{3}\)

\(x\)     \(=\frac{2}{6}\)rút gọn \(\frac{1}{3}\)

\(x-\frac{1}{4}=\frac{3}{5}\)

\(x=\frac{3}{5}+\frac{1}{4}\)

\(x=\frac{17}{20}\)

\(x:\frac{2}{3}=\frac{1}{2}\)

\(x=\frac{1}{2}.\frac{2}{3}\)

\(x=\frac{1}{3}\)

™Ta có: C=(x^2-1)(x^2-2)...(x^2-2006)

™Thay x=5 vào biểu thức C ta được :

™C=(5^2-1)(5^2-2)...(x^2-2016)=(25-1)(25-2)...(25-1016)=(25-1)(25-2)(25-3)(25-4)...(25-5)...(25-2016)=(25-1)(25-2)(25-3)(25-4)...(0)...(25-2016)

™Vì 0 nhân với số nào cũng bằng 0 nên C=0

™Vậy biểu thức C có giá trị bằng 0 tại x=5

--------------------------

™

(x^2-1)(x^2-2)...(x^2-2016)

để ý ta thấy C=(x^2-1)(x^2-2)...(x^2-25)...(x^2-2016)

thay x=5 vào ta có

C=(5^2-1)(5^2-2)...(5^2-25)...(5^2-2016)

C=(5^2-1)(5^2-2)....0...(5^2-2016)=0

 vậy C=0