chào bn mik đến từ năm 2022

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

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

\(\Rightarrow\left(x-1\right)^2.1-\left(x-1\right)^2.\left(x-1\right)^2=0\)

\(\Rightarrow\left(x-1\right)^2.\left[\left(x-1\right)^2-1\right]=0\)

\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^2=0\\\left(x-1\right)^2-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\\left(x-1-1\right)\left(x-1+1\right)=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x=2;x=0\end{cases}}}\)

Vậy: \(x\in\left\{1;2;0\right\}\)

(x - 1)² = (x - 1)⁴

<=> (x - 1)² - (x - 1)⁴ = 0

<=> (x - 1)² - (x - 1)².(x - 1)² = 0

<=> (x - 1)². [1 - (x - 1)²] = 0

<=> x - 1 = 0 hoặc 1 - (x - 1)² = 0

 <=>  x = 1             <=> (x - 1)² = 1

                               <=> x - 1 = 1 hoặc x - 1 = -1

                               <=> x = 2           <=> x = 0

Vậy x = 1; x = 2; x = 0

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

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

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

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

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

a)\(x\left(x-3\right)-2x+6=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)

b)\(\left(3x-5\right)\left(5x-7\right)+\left(5x+1\right)\left(2-3x\right)=4\)

\(\Leftrightarrow15x^2-46x+35-15x^2+7x+2-4=0\)

\(\Leftrightarrow33-39x=0\Leftrightarrow33=39x\Leftrightarrow x=\frac{33}{39}\)

a) \(x\left(x-3\right)-2x+6=0\)

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

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

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}}\)

b) \((3x-5)(5x-7)+(5x+1)(2-3x)=4\)

\(15x^2-46x+35+10x-15x^2+2-3x-4=0\)

\(33-39x=0\)

\(3\left(11-13x\right)=0\)

\(11-13x=0\)

\(13x=11\)

\(x=\frac{11}{13}\)

lớp11

\(a)24\times(x-16)=12^2\\\Rightarrow 24\times(x-16)=144\\\Rightarrow x-16=144:24\\\Rightarrow x-16=6\\\Rightarrow x=6+16\\\Rightarrow x=22\\---\\b)(x^2-10):5=5\\\Rightarrow x^2-10=5\times5\\\Rightarrow x^2-10=25\\\Rightarrow x^2=25+10\\\Rightarrow x^2=35\\\Rightarrow x=\pm\sqrt{35}\\---\)

\(c)(5x+335):2=400\\\Rightarrow 5x+335=400\times2\\\Rightarrow 5x+335=800\\\Rightarrow 5x=800-335\\\Rightarrow 5x=465\\\Rightarrow x=465:5\\\Rightarrow x=93\\---\\d)63:(5x+4)=2^3-1\\\Rightarrow 63:(5x+4)=8-1\\\Rightarrow 63:(5x+4)=7\\\Rightarrow 5x+4=63:7\\\Rightarrow 5x+4=9\\\Rightarrow 5x=9-4\\\Rightarrow 5x=5\\\Rightarrow x=5:5\)

\(\Rightarrow x=1\)

\(Toru\)

\(\Delta'=16-\left(3m+1\right)\ge0\Rightarrow m\le5\)

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=-8\\x_1x_2=3m+1\end{matrix}\right.\)

Kết hợp điều kiện đề bài ta được: \(\left\{{}\begin{matrix}x_1+x_2=-8\\5x_1-x_2=2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x_1+x_2=-8\\6x_1=-6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_1=-1\\x_2=-7\end{matrix}\right.\)

Thế vào \(x_1x_2=3m+1\)

\(\Rightarrow\left(-1\right).\left(-7\right)=3m+1\)

\(\Rightarrow m=2\) (thỏa mãn)

4. ( 3^x+3 + 3^x+1 ) = 3240

3^x+3 + 3^x+1 = 810

3^x . 3³ + 3^x . 3= 810

3^x. 30=810

3^x = 27

3^x = 3³ 

x=3

vậy x =3

4(3𝑥+3+3𝑥+1)=3240

4(3x+{\color{#c92786}{3}}+3x+{\color{#c92786}{1}})=32404(3x+3+3x+1)=3240

4(3𝑥+4+3𝑥)=3240

Đáp án

𝑥=403/3

k cho mk nha

cảm ơn các

bn rất nhiều

Bài 1 :

a, \(A=x\left(x-6\right)+10\)

=x^2 - 6x + 10

=x^2 - 2.3x+9+1

=(x-3)^2 +1 >0 Với mọi x dương

Cảm ơn bạn Vũ Anh Quân ;) ;) ;)