T nghĩ là hè năm nay á

\(23\%=\frac{23}{100}=0,23\)

\(5\%=\frac{5}{100}=\frac{1}{20}=0,05\)

\(136\%=\frac{136}{100}=\frac{34}{25}=1,36\)

\(225\%=\frac{225}{100}=\frac94=2,25\)

\(144\%=\frac{144}{100}=\frac{36}{25}=1,44\)

\(90\%=\frac{90}{100}=\frac{9}{10}=0,9\)

\(12,24\%=\frac{12,24}{100}=\frac{153}{1250}=0,1224\)

\(36,78\%=\frac{36,78}{100}=\frac{1839}{5000}=0,3678\)

\(1024,2\%=\frac{1024,2}{100}=\frac{5121}{500}=10,242\)

Sai thì bảo mình

`a) x^3 - 9x^2 + 14x = 0`

\(\Rightarrow\) `x^3 - 7x^2 - 2x^2 + 14x = 0`

\(\Rightarrow\) `(x^3 - 2x^2) - (7x^2 - 14x) =0`

\(\Rightarrow\) `x^2.(x - 2) - 7x.(x - 2) =0`

\(\Rightarrow\) `(x^2 - 7x)(x-2)=0`

\(\Rightarrow\) `x.(x-7)(x-2)=0`

\(\Rightarrow\left[\begin{array}{l}x=0\\ x-7=0\\ x-2=0\end{array}\right.\) \(\Rightarrow\left[\begin{array}{l}x=0\\ x=0+7\\ x=0+2\end{array}\right.\) \(\Rightarrow\left[\begin{array}{l}x=0\\ x=7\\ x=2\end{array}\right.\)

Vậy \(x\in\left\lbrace0;7;2\right\rbrace\)

`3.x^3 - 5x^2 + 8x - 4 = 0`

\(\Rightarrow\) `x^3 - x^2 - 4x^2 + 4x + 4x - 4 =0`

\(\Rightarrow\) `x^2 . (x-1) - 4x(x-1) + 4.(x-1) =0`

\(\Rightarrow\) `(x^2 - 4x + 4)(x-1)=0`

\(\Rightarrow\) `(x-2)^2(x-1)=0`

\(\Rightarrow\left[\begin{array}{l}x-2=0\\ x-1=0\end{array}\right.\) \(\Rightarrow\left[\begin{array}{l}x=2\\ x=1\end{array}\right.\)

Vậy \(x\in\left\lbrace2;1\right\rbrace\)

Đề có thiếu gì không bạn?

`1 + 2 + 3 + ... + 98+ 99 + 100`

Đặt `1 + 2 + 3 + ... + 98+ 99 + 100` là `A`

Số số hạng của `A` là :

`(100 - 1) : 1 + 1 = 100(` số hạng `)`

Tổng `A` là :

`(100 + 1)` x `100 : 2 = 5050`

Vậy ...

\(9+\left(-10\right)+11+\left(-12\right)+13+\left(-14\right)+15+\left(-16\right)\)

\(=\left\lbrack9+\left(-10\right)\right\rbrack+\left\lbrack11+\left(-12\right)\right\rbrack+\left\lbrack13+\left(-14\right)\right\rbrack+\left\lbrack15+\left(-16\right)\right\rbrack\)

\(=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)\)

\(=\left(-1\right).4=\left(-4\right)\)

\(x+\left(-5\right)=-18\)

=>x - 5 =-18

=> x = -18 + 5

=>x = -13

vậy x=-13

\(a)x^2-8x=x\left(x-8\right)\)

\(b\) \()x^2+16x=x\left(x+16\right)\)

\(c)3xy^2-4y\) \(=y.\left(3xy-4\right)\)

\(d)7x^2y+9x=x.\left(7xy+9\right)\)

gỡ cái gì á bạn?

\(A=\frac12+\frac{1}{2^2}+\frac{1}{2^3}+\cdots+\frac{1}{2^{100}}\)

Ta có:

\(2A=1+\frac12+\frac{1}{2^2}+\cdots+\frac{1}{2^{99}}\)

\(2A-A=\left(1+\frac12+\frac{1}{2^2}+\cdots+\frac{1}{2^{99}}\right)-\left(\frac12+\frac{1}{2^2}+\frac{1}{2^3}+\cdots+\frac{1}{2^{100}}\right)\)

\(A=1-\frac{1}{2^{100}}\)

Mà \(\frac{1}{2^{100}}<1\)

\(\rArr1-\frac{1}{2^{100}}<1\) hay \(A<1\)

Vậy \(A<1\)