bạn chép cả đề bài ra đi mình giải cho

Chứng minh tam giác NCM và tam giác MBA

Trả lời:

\(\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=\frac{2}{5}\\x+\frac{1}{2}=-\frac{2}{5}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{10}\\x=-\frac{9}{10}\end{cases}}\)

https://lazi.vn/users/dang_ky?u=kieu-anh.pham4

Trả lời:

\(\frac{4^6.9^9+6^9.102}{8^4.3^{12}-6^{11}}=\frac{\left(2^2\right)^6.\left(3^2\right)^6+\left(2.3\right)^9.2.3.17}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}=\frac{2^{12}.3^{12}+2^9.3^9.2.3.17}{2^{12}.3^{12}-2^{11}.3^{11}}\)

\(=\frac{2^{12}.3^{12}+2^{10}.3^{10}.17}{2^{12}.3^{12}-2^{11}.3^{11}}=\frac{2^{10}.3^{10}.\left(2^2.3^2+17\right)}{2^{10}.3^{10}.\left(2^2.3^2-2.3\right)}=\frac{2^2.3^2+17}{2^2.3^2-2.3}=\frac{53}{30}\)

\(3x+\frac{2}{2}x-1=5\)

\(\Rightarrow x\left(3+\frac{2}{2}\right)=5+1=6\) \(\Rightarrow4x=6\) \(\Rightarrow x=\frac{6}{4}=\frac{3}{2}=1,5\)

~~ HT ~~

\(3x+\frac{2}{2}\)\(x-1=5\)

\(\left(3+\frac{2}{2}\right)x=5+1\)

\(\left(3+\frac{2}{2}\right)x=6\)

\(\left(3+1\right)x=6\)

\(4x=6\)

\(x=\frac{6}{4}\)\(x=1,5\)

Hok tốt

(x-2)³=x-2

\(\Leftrightarrow\left(x-2\right)^2=1\)

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

Đây nhé bạn <3

(x - 2)³ = x - 2

=> (x - 2)³ - (x - 2) = 0

<=> (x - 2)[(x - 2)² - 1] = 0

<=> \(\orbr{\begin{cases}x-2=0\\\left(x-2\right)^2-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\\left(x-2\right)^2=1\end{cases}}\)

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

Vậy x \(\in\left\{1;2;3\right\}\)

Ta có A = |3 - x| + |5 + x| \(\ge\left|3-x+5+x\right|=\left|8\right|=8\)

Dấu "=" xảy ra <=> (3 - x )(5 + x) \(\ge0\)

Xét 2 trường hợp

TH1 : \(\hept{\begin{cases}3-x\ge0\\5+x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le3\\x\ge-5\end{cases}}\Leftrightarrow-5\le x\le3\)

TH2 : \(\hept{\begin{cases}3-x\le0\\5+x\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge3\\x\le-5\end{cases}}\Leftrightarrow x\in\varnothing\)

Vậy Min A = 8 <=> -5 \(\le x\le3\)