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a) 4x(x - 5) - (x - 1)(4x - 3) = 5
4x2 - 20x - (4x2 - 3x - 4x + 3) = 5
4x2 - 20x - 4x2 + 3x + 4x - 3 = 5
-13x - 3 = 5
\(\Rightarrow\) -13x = 8
\(\Rightarrow\) x = \(\dfrac{-8}{13}\)
b) (3x - 4)(x - 2) = 3x(x - 9) - 3
3x2 - 6x - 4x + 8 = 3x2 - 27x - 3
3x2 - 10x + 8 - 3x2 + 27x + 3 = 0
17x + 11 = 0
\(\Rightarrow\) 17x = -11
\(\Rightarrow\) x = \(\dfrac{-11}{17}\)
c) x2 - 81 = 0
\(\Rightarrow\) x2 = 81
\(\Rightarrow\) x = \(\pm\) 9
d) 3x2 - 75 = 0
3(x2 - 25) = 0
\(\Rightarrow\) x2 - 25 = 0
\(\Rightarrow\) x2 = 25
\(\Rightarrow\) x = \(\pm\)5
e) x2 - 4x + 3 = 0
x2 - x - 3x + 3 = 0
(x2 - x) - (3x - 3) = 0
x(x - 1) - 3(x - 1) = 0
(x - 3)(x - 1) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
xin lỗi vì chữa đề
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a: Ta có: \(4x\left(x-7\right)-4x^2=56\)
\(\Leftrightarrow4x^2-7x-4x^2=56\)
hay x=-8
b: Ta có: \(12x\left(3x-2\right)-\left(4-6x\right)=0\)
\(\Leftrightarrow36x^2-24x-4+6x=0\)
\(\Leftrightarrow36x^2-18x-4=0\)
\(\text{Δ}=\left(-18\right)^2-4\cdot36\cdot\left(-4\right)=900\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{18-30}{72}=\dfrac{-1}{6}\\x_2=\dfrac{18+30}{72}=\dfrac{2}{3}\end{matrix}\right.\)
c: Ta có: \(4\left(x-5\right)-\left(x-5\right)^2=0\)
\(\Leftrightarrow\left(x-5\right)\left(4-x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=9\end{matrix}\right.\)
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\(a,x^3+3x^2+3x=0\)
\(\Leftrightarrow x\left(x^2+3x+3\right)=0\)
\(\Leftrightarrow x=0\) Vì \(x^2+3x+3>0\forall x\)
\(b,x^3-3x^2+3x=0\)
\(\Leftrightarrow x\left(x^2-3x+3\right)=0\)
\(\Leftrightarrow x=0\)
\(c,\) bạn làm tương tự nha
c, x^3 + 6x^2 + 12x = 0
=> x(x^2 + 6x + 12) = 0
=> x(x^2 + 6x + 9 + 3) = 0
=> x[(x + 3)^2 + 3) = 0
=> x = 0 hoặc (x + 3)^2 + 3 = 0
=> x = 0 hoặc (x + 3)^2 = -3 (loại vì (x+3)^2 > 0)
vậy x = 0
a, x^3 + 3x^2 + 3x = 0
=> x(x^2 + 3x + 3) = 0
=>x(x^2 + 3x + 2,25 + 0,75) = 0
=> x[(x + 1,5)^2 + 0,75)] = 0
=> x = 0 hoặc (x + 1,5)^2 + 0,75 = 0
=> x = 0 hoặc (x + 1,5)^2 = -0,75 (loại)
vậy x = 0
b, x^3 - 3x^2 + 3x = 0
=> x(x^2 - 3x + 3) = 0
=> x(x^2 - 3x + 2,25 + 0,75) = 0
=> x[(x - 1,5)^2 + 0,75] = 0
=> x = 0 hoặc (x-1,5)^2 + 0,75 = 0
=> x = 0 hoặc (x - 1,5)^2 = -0,75 (loại)
vậy x = 0
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\(a,\)\(3x\left(x+1\right)-2x\left(x+2\right)=1-x\)
\(\Leftrightarrow3x^2+3x-2x^2-4x=1-x\)
\(\Leftrightarrow x^2-1=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
\(b,\)\(\frac{1}{3}x^2-4x+2x\left(2-3x\right)=0\)
\(\Leftrightarrow\frac{1}{3}x^2-4x+4x-6x^2=0\)
\(\Leftrightarrow-\frac{17}{3}x^3=0\)
\(\Leftrightarrow x=0\)
\(x\left(3x-2\right)-5\left(2-3x\right)=0\)
\(x\left(3x-2\right)+5\left(3x-2\right)=0\)
\(\left(x+5\right)\left(3x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+5=0\\3x-2=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{2}{3}\end{matrix}\right.\)
\(x\left(3x-2\right)-5\left(2-3x\right)=0\\ \Leftrightarrow\left(3x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)
Vậy...