1 + 1 = 2

Vì 2 - 1 = 1

1 + 0 = 1 

Vì 1 - 0 = 0

đang cần chứng minh nha bn

x-1 / x =y- 1/ y <=>(x^2 -1)/x = (y^2 -1)/y   => (X^2   -    1) y = (y^2   -1)x       => (x-y)( xy +1) =0 => x = y  hoac y = -1/x

voi x= y    =>  2x = x^3  +1     => bam may tinh giai ra 3 nghiem

voi y = -1/x thay vao ta dc -2/x = x^3     +1        => pt vo nghiem   vay pt co 3 nghiem     , nho dat dkcho x,y # 0 nha

\(x+1=\left(3-2x\right)\left(9\cdot-\sqrt{3x-2}\right)\)

Đề như vậy đúng ko ạ???

\(3x^2+2x-1=0\)

\(\Rightarrow3x^2+3x-x-1=0\)

\(\Rightarrow3x.\left(x+1\right)-\left(x+1\right)=0\)

\(\Rightarrow\left(x+1\right).\left(3x-1\right)=0\)

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

Vậy \(x\in\left\{-1;\dfrac{1}{3}\right\}\)

Chúc bạn học tốt nha!!!

Em làm bài này không chắc lắm! Nếu sai thì em xin lỗi anh Hoàng nha! Chưa thấy ai làm em làm đó nha!!!

Bài làm:

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

Lời giải:

Theo nhị thức New-ton:

\((x+1)^{2n}=C^{0}_{2n}+C^{1}_{2n}x+C^2_{2n}x^2+...+C^{2n}_{2n}x^{2n}\)

\((x-1)^n=C^0_{2n}-C^1_{2n}x+C^2_{2n}x^2-.....-C^{2n-1}_{2n}x^{2n-1}+C^{2n}_{2n}x^{2n}\)

Trừ theo vế ta có:

\(\frac{(x+1)^{2n}-(x-1)^{2n}}{2}=C^1_{2n}x+C^3_{2n}x^3+...+C^{2n-1}_{2n}x^{2n-1}\)

\(\Rightarrow \int ^{1}_{0}\frac{(x+1)^{2n}-(x-1)^{2n}}{2}dx=\int ^{1}_{0}(C^1_{2n}x+C^3_{2n}x^3+...+C^{2n-1}_{2n}x^{2n-1})dx\)

Xét vế trái:

\(\text{VT}=\frac{1}{2}\int ^{1}_{0}(x+1)^{2n}d(x+1)-\frac{1}{2}\int ^{1}_{0}(x-1)^{2n}d(x-1)\)

\(=\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{1}{2}\left ( \frac{(x+1)^{2n+1}-(x-1)^{2n+1}}{2n+1} \right )=\frac{2^{2n}-1}{2n+1}\)

Xét vế phải:

\(\text{VP}=\left.\begin{matrix} 1\\ 0\end{matrix}\right|\left ( \frac{C^{1}_{2n}x^2}{2}+\frac{C^{3}_{2n}x^4}{4}+....+\frac{C^{2n-1}_{2n}x^{2n}}{2n} \right )=\frac{1}{2}C^{1}_{2n}+\frac{1}{4}C^3_{2n}+...+\frac{1}{2n}C^{2n-1}_{2n}\)

Vậy \(A=\frac{2^{2n}-1}{2n+1}\)

Lời giải:

Ta có:

\(\log_2(x+4)+2\log_4(x+2)=2\log_{\frac{1}{2}}\frac{1}{8}=6\)

\(\Leftrightarrow 2\log_4(x+4)+2\log_4(x+2)=6\)

\(\Leftrightarrow \log_4(x+4)+\log_4(x+2)=3\)

\(\Leftrightarrow \log_4[(x+2)(x+4)]=3\)

\(\Leftrightarrow (x+2)(x+4)=4^3=64\)

\(\Leftrightarrow x^2+6x-56=0\)

\(\Leftrightarrow x=-3\pm \sqrt{65}\)

Kết hợp với ĐKXĐ ta suy ra \(x=-3+\sqrt{65}\) là nghiệm của pt

bạn ơi mình hỏi tí, sao log\(^{\left(x+4\right)}_2=2log^{\left(x+4\right)}_4\)

Bạn xem lại đề

=0,000000783 nhé bn

Theo Viet: \(\left\{{}\begin{matrix}z_1+z_2=2i\\z_1z_2=-1-2i\end{matrix}\right.\)

\(\Rightarrow z_1^3+z_2^3=\left(z_1+z_2\right)\left(z_1^2+z_2^2-z_1z_2\right)=\left(z_1+z_2\right)\left(\left(z_1+z_2\right)^2-3z_1z_1\right)\)

\(=2i\left[\left(2i\right)^2-3\left(-1-2i\right)\right]=2i\left(6i-1\right)=-12-2i\)