a/ Ta có : \(49.x^2-4=0\)

\(\Rightarrow49x^2=4\)

\(\Rightarrow x^2=\frac{4}{49}\Rightarrow\orbr{\begin{cases}x=\frac{-2}{7}\\x=\frac{2}{7}\end{cases}}\)

b/ \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=11\)

\(\left(x+3\right)\left(x+3\right)-\left(x+2\right)\left(x-2\right)=11\)

\(\Rightarrow\left(x^2+2.3.x+3^2\right)-\left(x^2-2^2\right)=11\)

\(\Rightarrow x^2+6x+9-x^2+4=11\)

\(\Rightarrow6x+13=11\)

\(\Rightarrow6x=11-13\)

\(\Rightarrow x=\frac{-2}{6}=\frac{-1}{3}\)

c/ \(\left(2x+1\right)^2-\left(x-3\right)^2-3\left(x+5\right)\left(x-5\right)=5\)

\(\Rightarrow\left(2x+1\right)\left(2x+1\right)-\left(x-3\right)\left(x-3\right)-3\left[\left(x+5\right)\left(x-5\right)\right]=5\)

\(\Rightarrow\left(4x^2+2.2x+1\right)-\left(x^2-2.3x+9\right)-3\left(x^2-25\right)\)\(=5\)

\(\Rightarrow\left(4x^2+4x+1\right)-\left(x^2-6x+9\right)-\left(3x^2-75\right)=5\)

\(\Rightarrow4x^2+4x+1-x^2+6x-9-3x^2+75=5\)

\(\Rightarrow\left(4x^2-x^2-3x^2\right)+\left(4x+6x\right)+\left(1-9+75\right)=5\)

\(\Rightarrow10x+67=5\)

\(\Rightarrow10x=5-67=-62\)

\(\Rightarrow x=\frac{-62}{10}=\frac{-31}{5}\)

d/ \(\left(3x+1\right)\left(3x-1\right)=8\)

\(\Rightarrow9x^2-1=8\)

\(\Rightarrow9x^2=8+1=9\)

\(\Rightarrow x^2=\frac{9}{9}=1\Leftrightarrow\orbr{\begin{cases}x=-1\\x=1\end{cases}}\)

Ai đó bấm hộ mình cái nút đúng đi!

Ta có : 49x² - 4 = 0

=> 49x² = 4

=> x² = 196

=> x² = 14² ; (-14)²

=> x = 14 ; -14

a) 49 . x² - 4 = 0

49 . x² = 4

7 ² . x² = 2²

(7 . x)² = 2²

7 . x = 2

x = 2 : 7

x = \(\frac{2}{7}\)

a) \(x^2-36=0\)

\(\Leftrightarrow x^2=36\)

\(\Leftrightarrow x=\pm\sqrt{36}=\pm6\)

b) \(\left(3x-5\right)^2-\left(x+6\right)^2=0\)

\(\Leftrightarrow\left(3x-5-x-6\right)\left(3x-5+x+6\right)=0\)

\(\Leftrightarrow\left(2x-11\right)\left(4x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{11}{2}\\x=\frac{-1}{4}\end{cases}}\)

d) \(4x^3-36x=0\)

\(\Leftrightarrow4x\left(x^2-9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}4x=0\\x^2-9=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm3\end{cases}}}\)

Vậy...

no nooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooô

câu này biết mới vừa trao đổi bài với thầy xong nhưng ko biết đúng ko

a) \(49x^2-1=0\)

\(\Rightarrow\left(7x\right)^2-1^2=0\)

\(\Rightarrow\left(7x-1\right)\left(7x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}7x-1=0\\7x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{7}\\x=\frac{-1}{7}\end{cases}}\)

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

\(\Rightarrow4x^2-4x+1-\left(4x^2-12x+x-3\right)=-3\)

\(\Rightarrow4x^2-4x+1-4x^2+12x-x+3=-3\)

\(\Rightarrow\left(4x^2-4x^2\right)+\left(-4x+12x-x\right)+4=-3\)

\(\Rightarrow7x+4=-3\)

\(\Rightarrow x=-1\)

c) \(9x\left(x-1\right)-\left(3x-1\right)^2=11\)

\(\Rightarrow\left(9x^2-9x\right)-\left(9x^2-6x+1\right)=11\)

\(\Rightarrow9x^2-9x-9x^2+6x-1-11=0\)

\(\Rightarrow-3x-12=0\)

\(\Rightarrow x=-4\)

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

\(\Rightarrow\orbr{\begin{cases}\left(2x+3\right)^2=0\\\left(x-1\right)^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}2x=-3\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-3}{2}\\x=1\end{cases}}\)

Bài 1.

\(a, (3x-4)^2\)

\(=\left(3x\right)^2-2\cdot3x\cdot4+4^2\)

\(=9x^2-24x+16\)

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

\(=1^2+2\cdot1\cdot4x+\left(4x\right)^2\)

\(=16x^2+8x+1\)

\(c,\left(2x+3\right)^3\)

\(=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot3+3\cdot2x\cdot3^2+3^3\)

\(=8x^3+36x^2+54x+27\)

\(d,\left(5-2x\right)^3\)

\(=5^3-3\cdot5^2\cdot2x+3\cdot5\cdot\left(2x\right)^2-\left(2x\right)^3\)

\(=125-150x+60x^2-8x^3\)

\(e,49x^2-25\)

\(=\left(7x\right)^2-5^2\)

\(=\left(7x-5\right)\left(7x+5\right)\)

\(f,\dfrac{1}{25}-81y^2\)

\(=\left(\dfrac{1}{5}\right)^2-\left(9y\right)^2\)

\(=\left(\dfrac{1}{5}-9y\right)\left(\dfrac{1}{5}+9y\right)\)

Bài 2.

\(a,\left(x-5\right)^2-\left(x+7\right)\left(x-7\right)=8\)

\(\Rightarrow x^2-2\cdot x\cdot5+5^2-\left(x^2-7^2\right)=8\)

\(\Rightarrow x^2-10x+25-\left(x^2-49\right)=8\)

\(\Rightarrow x^2-10x+25-x^2+49=8\)

\(\Rightarrow\left(x^2-x^2\right)-10x=8-25-49\)

\(\Rightarrow-10x=-66\)

\(\Rightarrow x=\dfrac{33}{5}\)

\(b,\left(2x+5\right)^2-4\left(x+1\right)\left(x-1\right)=10\)

\(\Rightarrow\left(2x\right)^2+2\cdot2x\cdot5+5^2-4\left(x^2-1^2\right)=10\)

\(\Rightarrow4x^2+20x+25-4x^2+4=10\)

\(\Rightarrow\left(4x^2-4x^2\right)+20x=10-25-4\)

\(\Rightarrow20x=-19\)

\(\Rightarrow x=\dfrac{-19}{20}\)

#\(Toru\)

Bài 1

a) (3x - 4)²

= (3x)² - 2.3x.4 + 4²

= 9x² - 24x + 16

b) (1 + 4x)²

= 1² + 2.1.4x + (4x)²

= 1 + 8x + 16x²

c) (2x + 3)³

= (2x)³ + 3.(2x)².3 + 3.2x.3² + 3³

= 8x³ + 36x² + 54x + 27

d) (5 - 2x)³

= 5³ - 3.5².2x + 3.5.(2x)² - (2x)³

= 125 - 150x + 60x² - 8x³

e) 49x² - 25

= (7x)² - 5²

= (7x - 5)(7x + 5)

f) 1/25 - 81y²

= (1/5)² - (9y)²

= (1/5 - 9y)(1/5 + 9y)

Bài 3.

\(a,A=x^2-6x+19\)

\(=x^2-6x+9+10\)

\(=\left(x^2-2\cdot x\cdot3+3^2\right)+10\)

\(=\left(x-3\right)^2+10\)

Ta thấy: \(\left(x-3\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-3\right)^2+10\ge10\forall x\)

Dấu \("="\) xảy ra \(\Leftrightarrow x-3=0\Leftrightarrow x=3\)

Vậy: \(Min_A=10\) khi \(x=3\)

\(b,B=-x^2+8x-20\)

\(=-x^2+8x-16-4\)

\(=-\left(x^2-8x+16\right)-4\)

\(=-\left(x^2-2\cdot x\cdot4+4^2\right)-4\)

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

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

\(\Rightarrow-\left(x-4\right)^2\le0\forall x\)

\(\Rightarrow-\left(x-4\right)^2-4\le-4\forall x\)

Dấu \("="\) xảy ra \(\Leftrightarrow x-4=0\Leftrightarrow x=4\)

Vậy \(Max_B=-4\) khi \(x=4\)

\(c,C=4x^2+12x+100\)

\(=4x^2+12x+9+91\)

\(=\left[\left(2x\right)^2+2\cdot2x\cdot3+3^2\right]+91\)

\(=\left(2x+3\right)^2+91\)

Ta thấy: \(\left(2x+3\right)^2\ge0\forall x\)

\(\Rightarrow\left(2x+3\right)^2+91\ge91\forall x\)

Dấu \("="\) xảy ra \(\Leftrightarrow2x+3=0\Leftrightarrow x=-\dfrac{3}{2}\)

Vậy \(Min_C=91\) khi \(x=\dfrac{-3}{2}\)

\(d,D=25+4x-x^2\)

\(=-x^2+4x-4+29\)

\(=-\left(x^2-2\cdot x\cdot2+2^2\right)+29\)

\(=-\left(x-2\right)^2+29\)

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

\(\Rightarrow-\left(x-2\right)^2\le0\forall x\)

\(\Rightarrow-\left(x-2\right)^2+29\le29\forall x\)

Dấu \("="\) xảy ra \(\Leftrightarrow x-2=0\Leftrightarrow x=2\)

Vậy \(Max_D=29\) khi \(x=2\)

#\(Toru\)

tìm GTLN nha mik QuÊn

tao vả bay lồn mày ấy hiểu chưa toán lớp 1 mà như cái phương chình đại học vậy hả thằng lồn ngu này

giúp mk vs