\(f\left(x\right)=-3x^2+x-1+x^4-x^3-x^2+3x^4+2x^3\)

\(f\left(x\right)=\left(x^4+3x^4\right)-\left(x^3-2x^3\right)-\left(3x^2+x^2\right)+x-1\)

\(f\left(x\right)=4x^4+x^3-4x^2+x-1\)

\(g\left(x\right)=x^4+x^2-x^3+x-5+5x^3-x^2-3x^4\)

\(g\left(x\right)=\left(x^4-3x^4\right)+\left(5x^3-x^3\right)+\left(x^2-x^2\right)+x-5\)

\(g\left(x\right)=-2x^4+4x^3+x-5\)

`@` `\text {Ans}`

`\downarrow`

`a,`

\(f(x) -3x^2 + x - 1 + x^4 - x^3 - x^2 + 3x^4 + 2x^3\)

`= (x^4 +3x^4) + (-x^3 +2x^3) + (-3x^2 - x^2) + x - 1`

`= 4x^4 + x^3 -4x^2 + x -1`

\(g(x) = x^4 + x^2 - x^3 + x - 5 + 5x^3 - x^2 - 3x^4\)

`= (x^4-3x^4) + (-x^3+5x^3) + (x^2 - x^2) + x -5`

`= -2x^4 + 4x^3 +x - 5`

`(2^x+1)^2 =25`

`=> (2^x+1)^2 = (+-5)^2`

\(\Rightarrow\left[{}\begin{matrix}2^x+1=5\\2^x+1=-5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2^x=4\\2^x=-6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x\in\varnothing\end{matrix}\right.\)

\(\left(x+6\right)\left(5^x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+6=0\\5^x-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-6\\5^x=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=0\end{matrix}\right.\)

\(\left(x-3\right)^{2023}=x-3\)

\(\Rightarrow\left(x-3\right)^{2023}-\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right)\left[\left(x-3\right)^{2022}-1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\\left(x-3\right)^{2022}-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\\left(x-3\right)^{2022}=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x-3=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

`#3107.101107`

1.

`(2^x + 1)^2 = 25`

`=> (2^x + 1)^2 = (+-5)^2`

`=>`\(\left[{}\begin{matrix}2^x+1=5\\2^x+1=-5\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}2^x=4\\2^x=-6\left(\text{vô lý}\right)\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=2\\x\in\varnothing\end{matrix}\right.\)

Vậy, `x =2.`

2.

`(x + 6)(5x - 1) = 0`

`=>`\(\left[{}\begin{matrix}x+6=0\\5x-1=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=-6\\5x=1\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=-6\\x=\dfrac{1}{5}\end{matrix}\right.\)

Vậy, `x \in {-6; 1/5}`

3.

`2*3^(x + 3) + 3^(2 + x) = 891`

`=> 2* 3^x * 3^3 + 3^2 * 3^x = 891`

`=> 54*3^x + 9*3^x = 891`

`=> 3^x * (54 + 9) = 891`

`=> 3^x * 63 = 891`

`=> 3^x = 891 \div 63`

`=> 3^x = 891/63`

Bạn xem lại đề.

4.

`(x - 3)^2023 = x - 3`

`=> (x - 3)^2023 - (x - 3) = 0`

`=> (x - 3) * [ (x - 3)^2022 - 1] = 0`

`=>`\(\left[{}\begin{matrix}x-3=0\\\left(x-3\right)^{2022}-1=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=3\\\left(x-3\right)^{2022}=1\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=3\\\left(x-3\right)^{2022}=\left(\pm1\right)^{2022}\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=3\\x-3=1\\x-3=-1\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=3\\x=4\\x=2\end{matrix}\right.\)

Vậy, `x \in {2; 3; 4}.`

Đề bài bn ghi thek thì ai làm nổi cho bn :V ?

ko có biết

mng giúp em với tối em nộp bài rồi a

cức + điên= lan ngọc cức điên

lớp 8 sao giống lớp 6 bọn em vậy

\(5^2.n-3-2.5^2=5^2.3\)

\(\Rightarrow25n-3-2.25=25.3\)

\(\Rightarrow25.n-3-50=75\)

\(\Rightarrow25.n-53=75\)

\(\Rightarrow25.n=128\)

\(\Rightarrow n=\frac{128}{25}\)

\(5^2.n-3-2.5^2=5^2.3\)

\(\Rightarrow5^2.\left(n-2\right)-3=5^2.3\)

\(\Rightarrow\left(n-2\right)-3=3\)

\(\Rightarrow n-2=3+3\)

\(\Rightarrow n-2=6\)

\(\Rightarrow n=6+2\)

\(\Rightarrow n=8\)