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g) \(\left(2x-1\right)^2-\left(2x+4\right)^2=0\)
\(\Leftrightarrow\left(2x-1+2x+4\right)\left(2x-1-2x-4\right)=0\)
\(\Leftrightarrow-5\left(4x+3\right)=0\)
\(\Leftrightarrow4x+3=0\)
\(\Leftrightarrow4x=-3\)
\(\Leftrightarrow x=\frac{-3}{4}\)
Vậy tập nghiệm của pt là \(S=\left\{\frac{-3}{4}\right\}\)
h) \(\left(2x-3\right)\left(3x+1\right)-x\left(6x+10\right)=30\)
\(\Leftrightarrow3x\left(2x-3\right)+\left(2x-3\right)-6x^2-10x=30\)
\(\Leftrightarrow6x^2-9x+2x-3-6x^2-10x=30\)
\(\Leftrightarrow-9x+2x-3-10x=30\)
\(\Leftrightarrow-17x-3=30\)
\(\Leftrightarrow-17x=33\)
\(\Leftrightarrow x=\frac{-33}{17}\)
Vậy tập nghiệm của pt là \(S=\left\{\frac{-33}{17}\right\}\)
a, 15x3 - 15x = 0
15x(x2-1)=0
15x=0 hoặc x2-1=0 (tự tính nhoa)
b,3x2-6x+3=0
3(x2-2x+1)=0
x2 -2x+1=0:3=3
x2-2x=3-1=2
x(x-2)=0
x=0 hoặc x-2=0 (tự tính nhoa)
Bài làm
a) 15x3-15x=0
<=> 15x( x2 - 1 ) = 0
<=> \(\orbr{\begin{cases}15x=0\\x^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}}\)
Vậy x = { 0; + 1 }
b) 3x2 - 6x + 3 = 0
<=> 3( x2 - 2x + 1 ) = 0
<=> x2 - 2x + 1 = 0
<=> ( x - 1 )2 = 0
<=> x - 1 = 0
<=> x = 1
Vậy x = 1
c) 5(x - 1) - 3x(1 - x) = 0
<=> 5(x - 1) + 3x(x - 1) = 0
<=> (5 + 3x)(x - 1) = 0
<=> \(\orbr{\begin{cases}5+3x=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{3}\\x=1\end{cases}}}\)
Vậy x = { -5/3; 1 }
e) -7(x + 2) = 2x(x + 2)
<=> -7(x + 2 ) - 2x( x + 2 ) = 0
<=> (x + 2)(-7 - 2x) = 0
<=> \(\orbr{\begin{cases}x+2=0\\-7-2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{7}{2}\end{cases}}}\)
Vậy x = { -2; x = -7/2 }
f)(2x - 3)(3x + 5) = (x - 1)(3x + 5)
<=> (2x - 3)(3x + 5) - (x - 1)(3x + 5) = 0
<=> (3x + 5)(2x - 3 - x + 1) = 0
<=> (3x + 5)(x - 2) = 0
<=> \(\orbr{\begin{cases}3x+5=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{3}\\x=2\end{cases}}}\)
Vậy x = { -5/3; 2 }
1,=\(x^2-3x-2x^2+6x=-x^2+3x\)
2,=\(3x^2-x-5+15x=3x^2+14x-5\)
3,=\(5x+15-6x^2-6x=-6x^2-x+15\)
4,=\(4x^2+12x-x-3=4x^2+11x-3\)
5: =>(x+5)^3=0
=>x+5=0
=>x=-5
6: =>(2x-3)^2=0
=>2x-3=0
=>x=3/2
7: =>(x-6)(x-10)=0
=>x=10 hoặc x=6
8: \(\Leftrightarrow x^3-12x^2+48x-64=0\)
=>(x-4)^3=0
=>x-4=0
=>x=4
\(x^4-3x^3+4x^2-3x-1=0\)
\(\Leftrightarrow x^4+x^3+2x^3+2x^2+2x^2+2x+x+1=0\)
\(\Leftrightarrow x^3\left(x+1\right)+2x^2\left(x+1\right)+2x\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x^3+2x^2+2x+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x^3+2x^2+2x+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow(x^3+x^2+x^2+x+x+1)\left(x+1\right)=0\)
\(\Leftrightarrow[x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)]\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+x+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}(x+1)^2=0\\x^2+x+1=0\end{cases}}\Rightarrow\hept{\begin{cases}x+1=0\\\varnothing\end{cases}}\Rightarrow x=-1\)
\(\left(3x-2\right)\left(x+6\right)\left(x^2+5\right)=0\)
\(TH1:3x-2=0\Leftrightarrow3x=2\Leftrightarrow x=\frac{2}{3}\)
\(TH2:x+6=0\Leftrightarrow x=-6\)
\(TH3:x^2+5=0\Leftrightarrow x^2=5\Leftrightarrow x=\sqrt{5}\)( ns vô nghiệm cx ko sai nha )
\(\left(2x+5\right)^2=\left(3x-1\right)^2\)
\(2x+5=3x-1\)
\(2x-3x=-1-5\)
\(-1x=-6\)
\(x=6\)
\(a,-2x\left(2-3x\right)+3\left(-5+7x-6x^2\right)=-4\)
\(\Rightarrow-4x+6x^2-15+21x-18x^2=-4\)
\(\Rightarrow-12x^2+17x-11=0\)
\(\Rightarrow12x^2-17x+11=0\)
\(\Rightarrow9x^2-2.3.\frac{17}{6}x+\left(\frac{17}{6}\right)^2-\left(\frac{17}{6}\right)^2+11=0\)
\(\Rightarrow\left(3x-\frac{17}{6}\right)^2+\frac{107}{36}=0VN\)
Không có gt x thỏa mãn
\(b,-3x\left(-1+3x-4x^2\right)+6x^2\left(-2x+3\right)=0\)
\(\Rightarrow3x-9x^2+12x^3-12x^3+18x^2=0\)
\(\Rightarrow9x^2+3x=0\)
\(\Rightarrow3x\left(3x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x=0\\3x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\3x=-1\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=-\frac{1}{3}\end{cases}}}\)
Đây nhé! Tích giúp mình nha
Ta có: \(\dfrac{4x^4+3x^3}{-x^3}+\dfrac{15x^2+6x}{3x}=0\)
\(\Leftrightarrow-4x-3+5x+2=0\)
\(\Leftrightarrow x-1=0\)
hay x=1