1/ ( x-3) ²=16

\(\Rightarrow\left[{}\begin{matrix}x-3=4\\x-3=-4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)

2/ (3x-1)³=8

\(\Rightarrow3x-1=2\\ \Rightarrow3x=3\\ \Rightarrow x=1\)

3/ (x-11)³=-27

\(\Rightarrow x-11=-3\\ \Rightarrow x=8\)

phần 4 mình ko rõ đề

4) \(x^3-3x^2+3x-1=-64\)

\(\Rightarrow x^3-3x^2+3x+63=0\\ \Rightarrow\left(x^3+3x^2\right)-\left(6x^2+18x\right)+\left(21x+63\right)=0\\ \Rightarrow x^2\left(x+3\right)+6x\left(x+3\right)+21\left(x+3\right)=0\\ \Rightarrow\left(x+3\right)\left(x^2+6x+21\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+3=0\\x^2+6x+21=0\end{matrix}\right.\)

\(x+3=0\\ \Rightarrow x=-3\)

\(x^2+6x+21=0\\ \Rightarrow\left(x^2+6x+9\right)+12=0\\ \Rightarrow\left(x+3\right)^2+12=0\)

Vì \(\left(x+3\right)^2\ge0;12>0\Rightarrow\left(x+3\right)^2+12>0\Rightarrow x^2+6x+21vônghiệm\)

Vậy \(x=-3\)

a) (2x-1)³ = 27
(2x-1)³ = 9³
2x-1 = 9
2x = 9+1
2x = 10
x = 10:5
x = 2
Vậy x = 2

b) (2x-1)⁴ = 81
(2x-1)⁴ = (\(\pm\)3⁴)
2x-1 = \(\pm\)3
Trường hợp 1:
2x-1 = 3
2x = 3+1
2x = 4
x = 4:2
x = 2
Trường hợp 2:
2x-1 = -3
2x = -3+1
2x = -2
x = -2:2
x = -1
Vậy x \(\in[_{ }2;-1]\)
Vì không tìm thấy ngoặc nhọn nên mình dùng tạm ngoặc vuông nhé

À phần b) bạn sửa dòng (2x-1)⁴ = (\(\pm\)3⁴) thành (2x-1)⁴ = (\(\pm\)3)⁴ nhé
Mình vừa viết nhầm

a, (2x-1)³=8

2x-1=2

2x=2+1

2x=3

x=1,5

a, \(\left(2x-1\right)^3=8\)

\(\Leftrightarrow\left(2x-1\right)^3=2^3\)

\(\Leftrightarrow2x-1=2\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)

b, \(\left(x-1\right)^x+2=\left(x-1\right)^x+4\)

\(\Leftrightarrow\left(x-1\right)^x+2-\left(x-1\right)^x-4=0\)

\(\Leftrightarrow-2\ne0\)=> vô nghiệm 

c, \(3^x-1+5.3^{3x}-1=162\)

Đề sai.

\(\frac{1}{3}.3^n+5.3^{n-1}=162\)

<=> \(3^{n-1}+5.3^{n-1}=162\)

<=> \(3^{n-1}\left(1+5\right)=162\)

<=> \(3^{n-1}.6=162\)

<=> \(3^{n-1}=162:6\)

<=> \(3^{n-1}=27\)

<=> \(3^{n-1}=3^3\)

<=> n - 1 = 3

<=> n = 3 + 1 = 4

Câu 1

a) Từ gt=>\(\hept{\begin{cases}x-5=1-3x\\x-5=3x-1\end{cases}}\)

<=>\(\hept{\begin{cases}4x=6\\2x=-4\end{cases}}\)

<=>\(\hept{\begin{cases}x=\frac{3}{2}\\x=-2\end{cases}}\)

b) Ta có: \(\hept{\begin{cases}\left(3x-1\right)^{100}\ge0,\forall x\in R\\\left(2y+1\right)^{200}\ge0,\forall x\in R\end{cases}}\)

Kết hợp với đề bài => \(\hept{\begin{cases}3x-1=0\\2y+1=0\end{cases}}\)

=>\(\hept{\begin{cases}x=\frac{1}{3}\\y=-\frac{1}{2}\end{cases}}\)

Bài 2

\(\frac{1}{3}.3^n+5.3^{n-1}=162\)

<=>\(3^{n-1}+5.3^{n-1}=162\)

<=>\(6.3^{n-1}=162\)

<=>\(3^{n-1}=27=3^3\)

<=>\(n-1=3\)

<=>\(n=4\)

b, Dấu = khi \(\hept{\begin{cases}x=\frac{1}{3}\\y=-\frac{1}{2}\end{cases}}\)

\(pt< =>6.3^{n-1}=162< =>3^{n-1}=3^3< =>n=4\)

d/

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)-\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-2\right)=0\)

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

e/

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

\(\Leftrightarrow x\left(x-1\right)-3\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)

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

f/

\(\Leftrightarrow3x\left(x-4\right)+12\left(x-4\right)=0\)

\(\Leftrightarrow3\left(x+4\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

g/

\(\Leftrightarrow\left(2x-1\right)\left(5-3x\right)-\left(x+2\right)\left(5-3x\right)=0\)

\(\Leftrightarrow\left(5-3x\right)\left(2x-1-x-2\right)=0\)

\(\Leftrightarrow\left(5-3x\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{5}{3}\end{matrix}\right.\)

a/

Đặt \(x-1=t\)

\(\Rightarrow\left(t+3\right)^4+\left(t-3\right)^4=162\)

\(\Leftrightarrow2t^4+108t^2=0\)

\(\Leftrightarrow2t^2\left(t^2+54\right)=0\)

\(\Leftrightarrow t=0\Rightarrow x=1\)

b/

Bạn coi lại đề, bài này ít nhất phải lớp 9 vì nghiệm xấu

c/

\(\Leftrightarrow x^2-\left(x^2+2x+1\right)=0\)

\(\Leftrightarrow-2x-1=0\)

\(\Leftrightarrow x=-\frac{1}{2}\)

3x+2\(\sqrt{162+n}\)+5(n+3)=0

ĐKXĐ: n \(\ge\) -162

<=>3x=-2\(\sqrt{162+n}\)-5(n+3)

x<-3n-9

=>3x<-9n-27

=>-9n-27>-2\(\sqrt{162+n}\)-5(n+3)

<=>9n+27>2\(\sqrt{162+n}\)+5(n+3)

<=>4n+12>2\(\sqrt{162+n}\)

<=>2n+6>\(\sqrt{162+n}\)

ĐK có nghiệm: n\(\ge\)-3

<=>4n²+24n+36>162+n

<=>4n²+23n-126>0

<=>\(\dfrac{-23+\sqrt{2545}}{8}< n\)hoặc n<\(\dfrac{-23-\sqrt{2545}}{8}\)

Vậy...

b: Ta có: \(x+3x-2=10\)

\(\Leftrightarrow4x=12\)

hay x=3