a) (3x + 4)² - (3x - 1).(3x + 1) = 49

=> (3x + 4).3x + (3x + 4).4 - (9x² - 1) = 49

=> 9x² + 12x + 12x + 16 - 9x² + 1 = 49

=> 24x + 17 = 49

=> 24x = 49 - 17

=> 24x = 32

=> \(x=\frac{32}{24}=\frac{4}{3}\)

Vậy \(x=\frac{4}{3}\)

b) (2x + 1)² - (x - 1)² = 0

=> (2x + 1 - x + 1).(2x + 1 + x - 1) = 0

=> (x + 2).3x = 0

=> (x + 2).x = 0

\(\Rightarrow\left[\begin{array}{nghiempt}x+2=0\\x=0\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}x=-2\\x=0\end{array}\right.\)

Vậy \(\left[\begin{array}{nghiempt}x=-2\\x=0\end{array}\right.\)

a)

➜9x²+24x+16-9x²+1=49

➜24x +17=49

⇒24x=49

⇒x=4/3

b)⇒4x²+4x+1-x²+2x-1=0

⇒3x²+6x=0

⇒3x(x+3)=0

⇒3x=0 hoặc x+3 =0

⇒x=0 hoặc x=-3

Bài 2: 

a: Ta có: \(x\left(2x-1\right)-2x+1=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)

\(c.\:\left(3x+4\right)^2-\left(3x+1\right)\left(3x-1\right)\\ =9x^2+24x+16-9x^2+1\\ 40x=-1\\ x=-\dfrac{1}{40}\)

\(d.\:\left(3x-1\right)^2-\left(3x-2\right)^2=0\\ \left(3x-1+3x-2\right)\left(3x-1-3x+2\right)=0\\ \left(6x-3\right)=0\\ x=\dfrac{1}{2}\)

\(g.\:\left(2x+1\right)^2-\left(x-1\right)^2=0\\ \left(2x+1+x-1\right)\left(2x+1-x+1\right)=0\\ 3x\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

c,\(\left(3x+4\right)^2-\left(3x-1\right)\left(3x+1\right)=49\)

\(\Rightarrow9x^2+24x+16-\left(9x^2-1\right)=49\)

\(\Rightarrow9x^2+24x+16-9x^2+1=49\)

\(\Rightarrow24x=49-1-16\)

\(\Rightarrow24x=32\Rightarrow x=\dfrac{4}{3}\)

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

\(\Rightarrow\left(3x-1-3x+2\right).\left(3x-1+3x-2\right)=0\)

\(\Rightarrow6x-3=0\Rightarrow6x=3\Rightarrow x=\dfrac{1}{2}\)

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

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

\(\Rightarrow\left(x+2\right).3x=0\Rightarrow x.\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

Chúc bạn học tốt!!!

a, 16x² - (4x - 5)² = 15

16x² - 4x² - 5² = 15

12x² - 5² = 15

12x² - 25 = 15

2x² = 15 + 25 = 40

x² = 40 : 2 

x² = 20

=> \(x=\sqrt{20}\)

Các câu kia tương tự

a: \(x\in\left\{0;25\right\}\)

c: \(x\in\left\{0;5\right\}\)

b: \(3x^2-2x-1=0\)

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

=>\(\left(x-1\right)\left(3x+1\right)=0\)

=>\(\left[{}\begin{matrix}x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)

a: Bạn ghi lại đề đi bạn

a)  \(x^3+3x^2+3x+2=0\)

<=>  \(x^3+x^2+x+2x^2+2x+2=0\)

<=>  \(x\left(x^2+x+1\right)+2\left(x^2+x+1\right)=0\)

<=>  \(\left(x+2\right)\left(x^2+x+1\right)=0\)

tự làm

b) \(x^4-2x^3+2x-1=0\)

<=>  \(\left(x^4-3x^3+3x^2-x\right)+\left(x^3-3x^2+3x-1\right)=0\)

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

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

<=>  \(\left(x-1\right)^3\left(x+1\right)=0\)

tự làm

c)   \(x^4-3x^3-6x^2+8x=0\)

<=>   \(x\left(x^3-3x^2-6x+8\right)=0\)

<=>  \(x\left[\left(x^3+x^2-2x\right)-\left(4x^2+4x-8\right)\right]=0\)

<=>\(x\left[x\left(x^2+x-2\right)-4\left(x^2+x-2\right)\right]=0\)

<=>   \(x\left(x-4\right)\left(x^2+x-2\right)=0\)

<=> \(x\left(x-4\right)\left(x-1\right)\left(x+2\right)=0\)

tự làm

Bài 1.

\( a)\dfrac{{4x - 8}}{{2{x^2} + 1}} = 0 (x \in \mathbb{R})\\ \Leftrightarrow 4x - 8 = 0\\ \Leftrightarrow 4x = 8\\ \Leftrightarrow x = 2\left( {tm} \right)\\ b)\dfrac{{{x^2} - x - 6}}{{x - 3}} = 0\left( {x \ne 3} \right)\\ \Leftrightarrow \dfrac{{{x^2} + 2x - 3x - 6}}{{x - 3}} = 0\\ \Leftrightarrow \dfrac{{x\left( {x + 2} \right) - 3\left( {x + 2} \right)}}{{x - 3}} = 0\\ \Leftrightarrow \dfrac{{\left( {x + 2} \right)\left( {x - 3} \right)}}{{x - 3}} = 0\\ \Leftrightarrow x - 2 = 0\\ \Leftrightarrow x = 2\left( {tm} \right) \)

Bài 2.

\(c)\dfrac{{x + 5}}{{3x - 6}} - \dfrac{1}{2} = \dfrac{{2x - 3}}{{2x - 4}}\)

ĐK: \(x\ne2\)

\( Pt \Leftrightarrow \dfrac{{x + 5}}{{3x - 6}} - \dfrac{{2x - 3}}{{2x - 4}} = \dfrac{1}{2}\\ \Leftrightarrow \dfrac{{x + 5}}{{3\left( {x - 2} \right)}} - \dfrac{{2x - 3}}{{2\left( {x - 2} \right)}} = \dfrac{1}{2}\\ \Leftrightarrow \dfrac{{2\left( {x + 5} \right) - 3\left( {2x - 3} \right)}}{{6\left( {x - 2} \right)}} = \dfrac{1}{2}\\ \Leftrightarrow \dfrac{{ - 4x + 19}}{{6\left( {x - 2} \right)}} = \dfrac{1}{2}\\ \Leftrightarrow 2\left( { - 4x + 19} \right) = 6\left( {x - 2} \right)\\ \Leftrightarrow - 8x + 38 = 6x - 12\\ \Leftrightarrow - 14x = - 50\\ \Leftrightarrow x = \dfrac{{27}}{5}\left( {tm} \right)\\ d)\dfrac{{12}}{{1 - 9{x^2}}} = \dfrac{{1 - 3x}}{{1 + 3x}} - \dfrac{{1 + 3x}}{{1 - 3x}} \)

ĐK: \(x \ne -\dfrac{1}{3};x \ne \dfrac{1}{3}\)

\( Pt \Leftrightarrow \dfrac{{12}}{{1 - 9{x^2}}} - \dfrac{{1 - 3x}}{{1 + 3x}} - \dfrac{{1 + 3x}}{{1 - 3x}} = 0\\ \Leftrightarrow \dfrac{{12}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} - \dfrac{{1 - 3x}}{{1 + 3x}} - \dfrac{{1 + 3x}}{{1 - 3x}} = 0\\ \Leftrightarrow \dfrac{{12 - {{\left( {1 - 3x} \right)}^2} - {{\left( {1 + 3x} \right)}^2}}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} = 0\\ \Leftrightarrow \dfrac{{12 + 12x}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} = 0\\ \Leftrightarrow 12 + 12x = 0\\ \Leftrightarrow 12x = - 12\\ \Leftrightarrow x = - 1\left( {tm} \right) \)

Bài 2.

\(a)5 + \dfrac{{96}}{{{x^2} - 16}} = \dfrac{{2x - 1}}{{x + 4}} - \dfrac{{3x - 1}}{{4 - x}}\)

ĐK: \(x\ne\pm4\)

\( Pt \Leftrightarrow \dfrac{{96}}{{\left( {x - 4} \right)\left( {x + 4} \right)}} - \dfrac{{2x - 1}}{{x + 4}} - \dfrac{{3x - 1}}{{x - 4}} = - 5\\ \Leftrightarrow \dfrac{{96 - \left( {2x - 1} \right)\left( {x - 4} \right) - \left( {3x - 1} \right)\left( {x + 4} \right)}}{{\left( {x - 4} \right)\left( {x + 4} \right)}} = - 5\\ \Leftrightarrow \dfrac{{ - 5{x^2} - 2x + 96}}{{\left( {x - 4} \right)\left( {x + 4} \right)}} = - 5\\ \Leftrightarrow - 5{x^2} - 2x + 96 = - 5\left( {{x^2} - 16} \right)\\ \Leftrightarrow 96 - 2x = 80\\ \Leftrightarrow - 2x = - 16\\ \Leftrightarrow x = 8\left( {tm} \right)\\ b)\dfrac{{3x + 2}}{{3x - 2}} - \dfrac{6}{{2 + 3x}} = \dfrac{{9{x^2}}}{{9{x^2} - 4}} \)

ĐK: \(x \ne \dfrac{2}{3};x \ne -\dfrac{2}{3}\)

\( Pt \Leftrightarrow \dfrac{{3x + 2}}{{3x - 2}} - \dfrac{6}{{2 + 3x}} - \dfrac{{9{x^2}}}{{9{x^2} - 4}} = 0\\ \Leftrightarrow \dfrac{{{{\left( {2 + 3x} \right)}^2} - 6\left( {3x - 2} \right) - 9{x^2}}}{{\left( {3x - 2} \right)\left( {2 + 3x} \right)}} = 0\\ \Leftrightarrow \dfrac{{16 - 6x}}{{\left( {3 - 2x} \right)\left( {2 + 3x} \right)}} = 0\\ \Leftrightarrow 16 - 6x = 0\\ \Leftrightarrow - 6x = - 16\\ \Leftrightarrow x = \dfrac{8}{3}\left( {tm} \right)\\ c)\dfrac{{x + 1}}{{{x^2} + x + 1}} - \dfrac{{x - 1}}{{{x^2} - x + 1}} = \dfrac{3}{{x\left( {{x^4} + {x^2} + 1} \right)}} \)

Ta có: \(x(x^4+x^2+1)=x[(x^2+1)^2-x^2]=x(x^2+x+1)(x^2-x+1)\)

Do \(\left\{ \begin{array}{l} {x^2} + x + 1 = {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} > 0\forall x\\ {x^2} - x + 1 = \left( {x - \dfrac{1}{2}} \right) + \dfrac{3}{4} > 0\forall x \end{array} \right.\) nên phương trình xác định với mọi $x \ne 0$

Quy đồng, rồi biến đổi phương trình về dạng \(2x=3 \Leftrightarrow x =\dfrac{3}{2} (tm)\)