a: \(5^{\left(x-2\right)\left(x+3\right)}=1\)

=>\(\left(x-2\right)\left(x+3\right)=0\)

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

c: \(\left|x^2+2x\right|+\left|y^2-9\right|=0\)

mà \(\left\{{}\begin{matrix}\left|x^2+2x\right|>=0\forall x\\\left|y^2-9\right|>=0\forall y\end{matrix}\right.\)

nên \(\left\{{}\begin{matrix}x^2+2x=0\\y^2-9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\left(x+2\right)=0\\\left(y-3\right)\left(y+3\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\in\left\{0;-2\right\}\\y\in\left\{3;-3\right\}\end{matrix}\right.\)

d: \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=120\)

=>\(2^x\left(1+2+2^2+2^3\right)=120\)

=>\(2^x\cdot15=120\)

=>\(2^x=8\)

=>x=3

e: \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)

=>\(\left(x-7\right)^{x+11}-\left(x-7\right)^{x+1}=0\)

=>\(\left(x-7\right)^{x+1}\left[\left(x-7\right)^{10}-1\right]=0\)

=>\(\left[{}\begin{matrix}x-7=0\\x-7=1\\x-7=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\\x=6\end{matrix}\right.\)

a/

Ta có

\(DN\perp HA\left(gt\right);BC\perp HA\left(gt\right)\) => DN//BC

\(\Rightarrow\widehat{NDB}+\widehat{CBD}=180^o\) (Hai góc trong cùng phía bù nhau)

\(\Rightarrow\widehat{NDA}+\widehat{ADB}+\widehat{ABD}+\widehat{ABC}=180^o\)

Ta có

tg ABD vuông cân tại A \(\Rightarrow\widehat{ADB}=\widehat{ABD}=45^o\Rightarrow\widehat{ADB}+\widehat{ABD}=90^o\)

\(\Rightarrow\widehat{NDA}+\widehat{ABC}=180^o-90^o=90^o\)

Xét tg vuông ABH

\(\widehat{BAH}+\widehat{ABC}=90^o\)

\(\Rightarrow\widehat{NDA}=\widehat{BAH}\)

Xét tg vuông NDA và tg vuông BAH có

\(\widehat{NDA}=\widehat{BAH}\left(cmt\right)\)

AD=AB (cạnh bên tg cân)

=> tg NDA = tg BAH (Hai tg vuông có cạnh huyền và góc nhọn tương ứng bằng nhau)

=> DN = AH

C/m tương tự ta cũng có tg vuông MAE = tg vuông CHA => EM=AH

b/

Ta có

\(DN\perp HA\left(gt\right);EM\perp HA\left(gt\right)\) => DN//EM

Xét tg vuông DIN và tg vuông EIM có

DN=EM (cùng bằng AH)

\(\widehat{IDN}=\widehat{IEM}\) (góc so le trong)

=> tg DIN = tg EIM (Hai tg vuông có 1 cạnh góc vuông và góc nhọn tương ứng bằng nhau)

=> DI=IE

\(\dfrac{3}{4}+\dfrac{9}{4}\left(x-\dfrac{1}{3}\right)^2=\dfrac{13}{16}\)

=>\(\dfrac{9}{4}\left(x-\dfrac{1}{3}\right)^2=\dfrac{13}{16}-\dfrac{3}{4}=\dfrac{1}{16}\)

=>\(\left(x-\dfrac{1}{3}\right)^2=\dfrac{1}{16}:\dfrac{9}{4}=\dfrac{1}{16}\cdot\dfrac{4}{9}=\dfrac{1}{36}\)

=>\(\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{1}{6}\\x-\dfrac{1}{3}=-\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}+\dfrac{1}{3}=\dfrac{1}{2}\\x=\dfrac{1}{3}-\dfrac{1}{6}=\dfrac{1}{6}\end{matrix}\right.\)

`3/4 + 9/4 (x-1/3)^2 =13/16`

`=>  9/4 (x-1/3)^2 =13/16-3/4`

`=>  9/4 (x-1/3)^2 =1/16`

`=> (x-1/3)^2 =1/16:9/4`

`=> (x-1/3)^2 =1/16 xx 4/9`

`=> (x-1/3)^2 =1/36`

`=> (x-1/3)^2 = (+- 1/6)^2`

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{1}{6}\\x-\dfrac{1}{3}=-\dfrac{1}{6}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{1}{6}\end{matrix}\right.\)

Bạn chụp lại đề nha bạn, đề hơi mờ á

Qua O, kẻ tia Oz//mn//Ax

Oz//Ax

=>\(\widehat{zOA}+\widehat{xAO}=180^0\)(hai góc trong cùng phía)

=>\(\widehat{zOA}=180^0-\widehat{xAO}=180^0-110^0=70^0\)

Oz//mn

=>\(\widehat{zOB}=\widehat{nBO}\)(hai góc so le trong)

=>\(\widehat{zOB}=60^0\)

\(\widehat{AOB}=\widehat{zOA}+\widehat{zOB}=70^0+60^0=130^0\)

`#3107.101107`

`(x - 1/3)^3 = -8/27`

`=> (x - 1/3)^3 = (-2/3)^3`

`=> x - 1/3 = -2/3`

`=> x = -2/3 + 1/3`

`=> x = -1/3`

Vậy, `x = -1/3.`

\(\left(\dfrac{x-1}{3}\right)^3=-\dfrac{8}{27}\)

\(\Rightarrow\left(\dfrac{x-1}{3}\right)^3=\left(\dfrac{-2}{3}\right)^3\)

\(\Rightarrow\dfrac{x-1}{3}=\dfrac{-2}{3}\)

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

\(\Rightarrow x=-2+1\)

\(\Rightarrow x=-1\)

Vậy $x=-1$.

\(A=-\dfrac{1}{2.3}-\dfrac{1}{3.4}-\dfrac{1}{4.5}-...-\dfrac{1}{9.10}\)

\(\Rightarrow-A=\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+\dfrac{5-4}{4.5}+...+\dfrac{10-9}{9.10}=\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}=\)

\(=\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{2}{5}\Rightarrow A=-\dfrac{2}{5}\)

3:

\(A=\dfrac{1}{7}+\dfrac{1}{7^2}+...+\dfrac{1}{7^{100}}\)

=>\(7A=1+\dfrac{1}{7}+...+\dfrac{1}{7^{99}}\)

=>\(7A-A=1+\dfrac{1}{7}+...+\dfrac{1}{7^{99}}-\dfrac{1}{7}-\dfrac{1}{7^2}-...-\dfrac{1}{7^{100}}\)

=>\(6A=1-\dfrac{1}{7^{100}}=\dfrac{7^{100}-1}{7^{100}}\)

=>\(A=\dfrac{7^{100}-1}{7^{100}\cdot6}\)

2:

\(19M=\dfrac{19^{31}+95}{19^{31}+5}=1+\dfrac{90}{19^{31}+5}\)

\(19N=\dfrac{19^{32}+95}{19^{32}+5}=\dfrac{19^{32}+5+90}{19^{32}+5}=1+\dfrac{90}{19^{32}+5}\)

\(19^{31}+5< 19^{32}+5\)

=>\(\dfrac{90}{19^{31}+5}>\dfrac{90}{19^{32}+5}\)

=>19M>19N

=>M>N