Bài 6: Lũy thừa của một số hữu tỉ (tiếp theo...)

\(3\cdot(-2)^{5x+8}+5^2=1\\\Rightarrow 3\cdot(-2)^{5x+8}+25=1\\\Rightarrow 3\cdot(-2)^{5x+8}=1-25\\\Rightarrow 3\cdot(-2)^{5x+8}=-24\\\Rightarrow (-2)^{5x+8}=-24:3\\\Rightarrow (-2)^{5x+8}=-8\\\Rightarrow (-2)^{5x+8}=(-2)^3\\\Rightarrow 5x+8=3\\\Rightarrow 5x=3-8\\\Rightarrow 5x=-5\\\Rightarrow x=-5:5\\\Rightarrow x=-1\)

\(-\dfrac{2}{3^3}:0,\left(3\right)+\left(\dfrac{2}{3}\right)^2\)

\(=\dfrac{-2}{27}:\dfrac{1}{3}+\dfrac{4}{9}\)

\(=\dfrac{-2}{9}+\dfrac{4}{9}=\dfrac{2}{9}\)

`#3107.101107`

`(-2/3)^3 \div 0,(3) + (2/3)^2`

`= (-2/3)^3 \div 1/3 + (2/3)^2`

`= (-2/3)^3 * 3 + (2/3)^2`

`= (2/3)^2 * (-2/3 * 3 + 1)`

`= 4/9 * (-2+1)`

`= 4/9 * (-1)`

`= -4/9`

\(\sqrt{\dfrac{16}{9}}+\left(\dfrac{2}{3}\right)^9:\left(-\dfrac{2}{3}\right)^8-\left|-2023\right|\)

\(=\dfrac{4}{3}-2023+\left(\dfrac{2}{3}\right)^9:\left(\dfrac{2}{3}\right)^8\)

\(=\dfrac{4}{3}+\dfrac{2}{3}-2023\)

=2-2023

=-2021

\(2^x+11.3^4=12^x\)

\(2^x\)chẵn, \(11.3^4\) lẻ => \(2^x+11.3^4\) lẻ(1)

Mà \(12^x\) chẵn(2)

Từ (1) và (2) => \(VT\ne VP\)

=> không tồn tại x thỏa mã phương trình

Cách trên là với điều kiện \(x\in N\cdot\) nha, cách này là với trường hợp không có điều kiện của x

\(2^x+11.3^4=12^x=2^{2x}.3^x\)

\(2^x\left(6^x-1\right)=11.3^4\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2^x=11\\6^x-1=3^4\end{matrix}\right.\\\left\{{}\begin{matrix}6^x=12\\2^x=3^4\end{matrix}\right.\end{matrix}\right.\)

+) Nếu x=0 => Loại

+) Nếu \(x\in N^{\cdot}\)

-) \(2^x=11\) (Loại vì 2^x chãn)

-) \(6^x=12\Leftrightarrow2^x.3^x=2^2.3\)

\(\Leftrightarrow\left\{{}\begin{matrix}2^x=2^2\\3^x=3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=1\end{matrix}\right.\) (Loại)

+) Nếu x>0; \(x\notin Z\)

=> \(2^x;6^x\notin Z\) (Loại)

+) Nếu x<0 => \(\left\{{}\begin{matrix}2^x< 2\\6^x< 6\end{matrix}\right.\) (Loại)

=> Không tồn tại x thỏa mãn phương trình

\(\left(x-2\right)^4+\left(2y-1\right)^{2024}\le0\left(1\right)\)

Vì \(\left\{{}\begin{matrix}\left(x-2\right)^4\ge0\forall x\\\left(2y-1\right)^{2024}\ge0\forall x\end{matrix}\right.\)

\(\Rightarrow\left(x-2\right)^4+\left(2y-1\right)^{2024}\ge0\left(2\right)\)

Từ (1) và (2)

\(\Rightarrow\left(x-2\right)^4+\left(2y-1\right)^{2024}=0\)

\(\Rightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{1}{2}\end{matrix}\right.\)

\(M=21.2^2.\dfrac{1}{2}+4.2.\left(\dfrac{1}{2}\right)^2=21.2+4.2.\dfrac{1}{4}=42+2=44\)

Ta có: \(\left(x-2\right)^4\ge0\forall x\)

           \(\left(2y-1\right)^{2024}\ge0\forall y\)

\(\Rightarrow\left(x-2\right)^4+\left(2y-1\right)^{2024}\ge0\forall x;y\)

Mặt khác: \(\left(x-2\right)^4+\left(2y-1\right)^{2024}\le0\)

nên \(\left(x-2\right)^4+\left(2y-1\right)^{2024}=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)^4=0\\\left(2y-1\right)^{2024}=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\2y-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{1}{2}\end{matrix}\right.\)

Thay \(x=2\) và \(y=\dfrac{1}{2}\) vào \(M\), ta được:

\(M=21\cdot2^2\cdot\dfrac{1}{2}+4\cdot2\cdot\left(\dfrac{1}{2}\right)^2\)

\(=42+2\)

\(=44\)

Vậy \(M=44\) tại \(x=2;y=\dfrac{1}{2}\).

#\(Toru\)

Đặt `A=1/3+1/(3^2)+...+1/(3^100)`

`3A=1+1/3+...+1/(3^99)`

`3A-A=(1+1/3+...+1/(3^99))-(1/3+1/(3^2)+...+1/(3^100))`

`2A=1-1/(3^100)`

`A=(1-1/(3^100))/2`

Đặt \(A=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\)

\(3\cdot A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)

\(3A-A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\right)\)

\(2A=1-\dfrac{1}{3^{100}}\)

\(\Rightarrow A=\dfrac{3^{100}-1}{3^{100}}:2=\dfrac{3^{100}-1}{2\cdot3^{100}}\)

#\(Toru\)

\(\left(3x-5y\right)^2+\left(xy-135\right)^2=0\)

Vì \(\left\{{}\begin{matrix}\left(3x-5y\right)^2\ge0\forall x,y\\\left(xy-135\right)^2\ge0\forall x,y\end{matrix}\right.\)

=>  \(\left\{{}\begin{matrix}3x-5y=0\\xy-135=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}y\\xy=135\end{matrix}\right.\)

\(\Rightarrow\dfrac{5}{3}y.y=135\)\(\Rightarrow y^2=81\)

\(\Leftrightarrow\left[{}\begin{matrix}y=9\Rightarrow x=15\\y=-9\Rightarrow x=-15\end{matrix}\right.\)

Ta có: \(\left(3x-5y\right)^2\ge0\forall x;y\)

           \(\left(xy-135\right)^2\ge0\forall x;y\)

\(\Rightarrow\left(3x-5y\right)^2+\left(xy-135\right)^2\ge0\forall x\)

Mặt khác: \(\left(3x-5y\right)^2+\left(xy-135\right) ^2=0\)

nên: \(\left\{{}\begin{matrix}\left(3x-5y\right)^2=0\\\left(xy-135\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x-5y=0\\xy-135=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x=5y\\xy=135\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}y\\\dfrac{5}{3}y^2=135\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}y\\y^2=81\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}y\\\left[{}\begin{matrix}y=9\\y=-9\end{matrix}\right.\end{matrix}\right.\)

\(+,TH1:y=9\Leftrightarrow x=\dfrac{5}{3}\cdot9=15\left(tm\right)\)

\(+,TH2:y=-9\Leftrightarrow x=\dfrac{5}{3}\cdot\left(-9\right)=-15\left(tm\right)\)

Vậy ...

#\(Toru\)

`# \text {DNamNguyenV}`

`a,`

Ta có: M là trung điểm của BC

`=> \text {MB = MC}`

Xét `\Delta ABM` và `\Delta ECM`:

`\text {MA = ME (gt)}`

\(\text{ }\widehat{\text{ AMB}}=\widehat{\text{EMC}}\left(\text{2 góc đối đỉnh}\right)\)

`\text {MB = MC}`

`=> \Delta ABM = \Delta ECM (c - g - c)`

`b,`

Vì `\Delta ABM = \Delta ECM (a)`

`=> \text {AB = CE (2 góc tương ứng)}`

a: Xét ΔBAD và ΔBED có

BA=BE

góc ABD=góc EBD

BD chung

Do đó: ΔBAD=ΔBED

b: ΔBAD=ΔBED

=>DA=DE

`#040911`

`A = 1 + 2 + 4 + 8 + ... + 8192`

`A = 1 + 2 + 2^2 + 2^3 + ... + 2^13`

`2A = 2 + 2^2 + 2^3 + 2^4 + ... + 2^14`

`2A - A = (2 + 2^2 + 2^3 + ... + 2^14) - (1 + 2 + 2^2 + ... + 2^13)`

`A = 2 + 2^2 + 2^3 + ... + 2^14 - 1 - 2 - 2^2 - ... - 2^13`

`A = 2^14 - 1`

Vậy, `A = 2^14 - 1.`

A=1+2+2^2+...+2^13

=>2A=2+2^2+2^3+...+2^14

=>2A-A=2^14-1

=>A=2^14-1