- Help Me :((

\(25\left(x+3\right)^2+\left(1-5x\right)\left(1+5x\right)=0\)

\(25\left(x^2+6x+9\right)+1-25x^2=0\)

\(25x^2+150x+225+1-25x^2=0\)

\(150x=-226\)

\(x=-\frac{113}{75}\)

25 ( x + 3 )² + ( 1 - 5x )( 1 + 5x ) = 0

25 ( x² + 6x + 9 ) + 1 + 5x - 5x - 25x² = 0

25x² + 150x + 225 + 1 + 5x - 5x - 25x² = 0

150x + 226 = 0

150x = -226

x = -226/150

a) Ta có :  5 x : 5 2 = 5 2  nên x = 4,

b) Ta có:  5 x + 1 : 5 = 5 4  nên x = 4.

c) Ta có :  4 2 x - 1 : 4 = 4 2  nên x = 2. 

1) \(\left(5x-1\right)\left(5x+1\right)=25x^2-7x+15\)

\(\Leftrightarrow25x^2-1=25x^2-7x+15\)

\(\Leftrightarrow7x=16\Leftrightarrow x=\dfrac{16}{7}\)

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

\(\Leftrightarrow3x^2-2x-5-3x^2-2x+1=x-4\)

\(\Leftrightarrow5x=0\Leftrightarrow x=0\)

a: Ta có: \(3x\left(3x-1\right)-\left(3x+1\right)\left(3x-1\right)=0\)

\(\Leftrightarrow9x^2-3x-9x^2+1=0\)

\(\Leftrightarrow3x=1\)

hay \(x=\dfrac{1}{3}\)

b: Ta có: \(x^2-5x+25-5x=0\)

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

\(\Leftrightarrow x-5=0\)

hay x=5

\(\dfrac{3}{5}x-\dfrac{25}{4}=\dfrac{1}{2}:x+\dfrac{5}{4}\)

=>\(\dfrac{3}{5}x-\dfrac{25}{4}-\dfrac{1}{2x}+\dfrac{5}{4}\)

=>\(\dfrac{3}{5}x-\dfrac{1}{2x}=\dfrac{30}{4}\)

=>\(\dfrac{3x}{5}-\dfrac{1}{2x}=\dfrac{30}{4}\)

=>\(\dfrac{6x^2-5}{10x}=\dfrac{30}{4}\)

=>\(6x^2-5=10x\cdot\dfrac{30}{4}=5x\cdot15=75x\)

=>\(6x^2-75x-5=0\)

=>\(x=\dfrac{75\pm\sqrt{5745}}{12}\)

Ừ nhưng thấy kêu kh tìm được số lớn. Bạn có cách giải khác kh?

C2:

Số số hạng của tổng là: [(x + 9) - (x + 1)]:2 + 1 = 5 (số)

Áp dụng cách tính tổng các số cách đều ta có:

[(x + 9) + (x + 1)].5 : 2 = \(\frac{5\left(2x+10\right)}{2}=0\)

=> 5(2x + 10) = 0

=> 2x + 10 = 0

=> 2x = -10

=> x = -5

a) Ta có :  2 x : 2 2 = 2 5  nên x = 7.

b) Ta có:  3 x : 3 2 = 3 5  nên x = 7.

c) Ta có :  4 4 : 4 x = 4 2  nên x = 2.

d) Ta có :  5 x : 5 2 = 5 2  nên x = 4,

e) Ta có:  5 x + 1 : 5 = 5 4  nên x = 4.

f) Ta có :  4 2 x - 1 : 4 = 4 2  nên x = 2

\(x=\dfrac{7}{25}+\dfrac{-1}{5}=\dfrac{7}{25}-\dfrac{1}{5}=\dfrac{2}{25}.\\ x=\dfrac{5}{11}+\dfrac{4}{-9}=\dfrac{5}{11}-\dfrac{4}{9}=\dfrac{1}{99}.\\ \dfrac{5}{9}-\dfrac{x}{-1}=\dfrac{-1}{3}\Leftrightarrow\dfrac{5}{9}+x=-\dfrac{1}{3}.\Leftrightarrow x=-\dfrac{8}{9}.\)

\(x=\dfrac{7}{25}+-\dfrac{1}{5}=>\dfrac{7}{25}+-\dfrac{5}{25}=>x=\dfrac{2}{25}\)

\(x=\dfrac{5}{11}+\dfrac{4}{-9}=>\dfrac{-45}{-99}+\dfrac{44}{-99}=>x=\dfrac{-1}{-99}=\dfrac{1}{99}\)

\(\dfrac{5}{9}-\dfrac{x}{-1}=-\dfrac{1}{3}=>-\dfrac{1}{3}-\dfrac{5}{9}=>\dfrac{x}{-1}=-\dfrac{8}{9}=>x=-\dfrac{8}{9}\)

\(x=\dfrac{7}{25}+\dfrac{-1}{5}\\ \Rightarrow x=\dfrac{7}{25}+\dfrac{-5}{25}\\ \Rightarrow x=\dfrac{2}{25}\\ x=\dfrac{5}{11}+\dfrac{4}{-9}\\ \Rightarrow x=\dfrac{-45}{-99}+\dfrac{44}{-99}\\ \Rightarrow x=\dfrac{45}{99}+\dfrac{44}{99}\\ \Rightarrow x=\dfrac{95}{99}\\ \dfrac{5}{9}-\dfrac{x}{-1}=\dfrac{-1}{3}\\ \Rightarrow\dfrac{x}{-1}=\dfrac{5}{9}-\dfrac{-1}{3}\\ \Rightarrow\dfrac{x}{-1}=\dfrac{5}{9}-\dfrac{-3}{9}\\ \Rightarrow\dfrac{x}{-1}=\dfrac{8}{9}\Rightarrow x\cdot9=-1\cdot8=-8\\ \Rightarrow x\cdot9=-8\\ \Rightarrow x=\dfrac{-8}{9}\)

a) Đặt: \(A=1+2^2+2^3+...+2^{10}\)

\(\Rightarrow2A=2\left(1+2^2+2^3+...+2^9+2^{10}\right)\)

\(\Rightarrow2A=2+2^3+2^4+...+2^{10}+2^{11}\)

\(\Rightarrow2A-A=\left(2+2^3+2^4+...+2^{10}+2^{11}\right)-\left(1+2^2+2^3+...+2^{10}\right)\)

\(\Rightarrow A=\left(2^3-2^3\right)+\left(2^4-2^4\right)+...+\left(2-1\right)+\left(2^{11}-2^2\right)\)

\(\Rightarrow A=0+0+...+1+\left(2^{11}-2^2\right)\)

\(\Rightarrow A=1+2^{11}-2^2=1+2048-4=2045\)

Vậy: \(1+2^2+2^3+...+2^{10}=2045\)

b) 

a] \(60-3\left(x-1\right)=2^3\cdot3\)

\(\Rightarrow60-3\left(x-1\right)=24\)

\(\Rightarrow3\left(x-1\right)=36\)

\(\Rightarrow x-1=12\)

\(\Rightarrow x=13\)

b] \(\left(3x-2\right)^3=2\cdot2^5\)

\(\Rightarrow\left(3x-2\right)^3=2^6\)

\(\Rightarrow\left(3x-2\right)^3=\left(2^2\right)^3\)

\(\Rightarrow3x-2=2^2\)

\(\Rightarrow3x=6\)

\(x=2\)

c] \(5^{x+1}-5^x=500\)

\(\Rightarrow5^x\left(5-1\right)=500\)

\(\Rightarrow5^x\cdot4=500\)

\(\Rightarrow5^x=125\)

\(\Rightarrow5^x=5^3\)

\(\Rightarrow x=3\)

d] \(x^2=x^4\)

\(\Rightarrow x=x^2\)

\(\Rightarrow x-x^2=0\)

\(\Rightarrow x\left(1-x\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

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