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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 \(\left[\begin{array}{nghiempt}x=-2\\x=0\end{array}\right.\)
\(c.\:\left(3x+4\right)^2-\left(3x+1\right)\left(3x-1\right)\\ =9x^2+24x+16-9x^2+1\\ 40x=-1\\ x=-\dfrac{1}{40}\)
\(d.\:\left(3x-1\right)^2-\left(3x-2\right)^2=0\\ \left(3x-1+3x-2\right)\left(3x-1-3x+2\right)=0\\ \left(6x-3\right)=0\\ x=\dfrac{1}{2}\)
\(g.\:\left(2x+1\right)^2-\left(x-1\right)^2=0\\ \left(2x+1+x-1\right)\left(2x+1-x+1\right)=0\\ 3x\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
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) \(x^3+3x^2+3x+2=0\)
<=> \(x^3+x^2+x+2x^2+2x+2=0\)
<=> \(x\left(x^2+x+1\right)+2\left(x^2+x+1\right)=0\)
<=> \(\left(x+2\right)\left(x^2+x+1\right)=0\)
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: \(x \ne -\dfrac{1}{3};x \ne \dfrac{1}{3}\)
\( Pt \Leftrightarrow \dfrac{{12}}{{1 - 9{x^2}}} - \dfrac{{1 - 3x}}{{1 + 3x}} - \dfrac{{1 + 3x}}{{1 - 3x}} = 0\\ \Leftrightarrow \dfrac{{12}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} - \dfrac{{1 - 3x}}{{1 + 3x}} - \dfrac{{1 + 3x}}{{1 - 3x}} = 0\\ \Leftrightarrow \dfrac{{12 - {{\left( {1 - 3x} \right)}^2} - {{\left( {1 + 3x} \right)}^2}}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} = 0\\ \Leftrightarrow \dfrac{{12 + 12x}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} = 0\\ \Leftrightarrow 12 + 12x = 0\\ \Leftrightarrow 12x = - 12\\ \Leftrightarrow x = - 1\left( {tm} \right) \)
lớp 8 đến bây giờ đã học hằng đẳng thức chưa ạ ? Để mình đưa bạn hướng giải câu đầu !
câu sau : x^2 -9^2 = 0<=> (x-9)^2=0 <=>x-9=0 <=> x=9
a, ( 3x - 1 )^2 - 3x( 3x + 2 ) = 0
<=>9x2-6x+1-9x2-6x=0
<=>-12x+1=0
<=>-12x=-1
<=>x=1/12
b, ( 2x + 3)^2 = 4x(x + 1 )
<=>(2x+3)2-4x(x+1)=0
<=>4x2+12x+9-4x2-4x=0
<=>8x+9=0
<=>8x=-9
<=>x=-9/8
c) vô fx gõ lại
d)x2-4x+4=16
<=>(x-2)2-16=0
<=>(x-2)2-42=0
<=>(x-2+4)(x-2-4)=0
<=>(x+2)(x-6)=0
<=>x+2=0 hoặc x-6=0
<=>x=-2 hoặc x=6
c(x-1)^2=4
x^2-2x+1=4
x^2-2x+1-4=0
x^2-2x-3=0
x^2-3x+x-3=0
x(x-3)+(x-3)=0
(x-3)(x+1)=0
\(\Rightarrow\hept{\begin{cases}x-3=0\\x+1=0\end{cases}\Rightarrow\hept{\begin{cases}x=3\\x=-1\end{cases}}}\)
d, x^3+2x^2-x-2=0
x^2(x+2)-(x+2)=0
(x+2)(x^2-1)=0
\(\Rightarrow\hept{\begin{cases}x=-2\\x=+-1\end{cases}}\)
(3x+4)\(^2\) - (3x-1)(3x+1)=49
=>\(9\text{x}^2+24x+16-9\text{x}^2+1\)\(=49\)
=>\(24\text{x}+17=49\)
=> 24x = 32
=> x = \(\dfrac{4}{3}\)
b) \(\left(3\text{x}-1\right)^2-\left(3\text{x}-2\right)^2=0 \)
\(=>9\text{x}^2-6\text{x}+1-9\text{x}^2+12\text{x}-4=0\)
\(=>6\text{x}-3=0\)
=> 6x = 3
=> x = \(\dfrac{1}{2}\)
c) \(\left(2\text{x}+1\right)^2-\left(x-1\right)^2=0\)
\(=>4\text{x}^2+4\text{x}+1-x^2+2\text{x}-1=0\)
=> \(3\text{x}^2+6\text{x}=0\)
=> \(3\text{x}\left(x+2\right)=0\)
=> 3x=0 hoặc x+2 = 0
+) 3x = 0 => x =0
+) x+2 = 0 => x = -2