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a)
\(\left(5x+3\right)\cdot\left(x^2+4\right)\cdot\left(x-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}5x+3=0\\x-4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\frac{3}{5}\\x=4\end{matrix}\right.\)
b)
\(\left(4x-1\right)\cdot\left(x-3\right)-\left(x-2\right)\cdot\left(5x+2\right)=0\\ \Leftrightarrow4x^2-12x-x+3-5x^2-2x+10x+4=0\\ \Leftrightarrow-x^2-5x+7=0\\ \Rightarrow x=\left[{}\begin{matrix}-\frac{5+\sqrt{53}}{2}\\-\frac{5-\sqrt{53}}{2}\end{matrix}\right.\)
c)
\(\left(x+3\right)\cdot\left(x-5\right)+\left(x+3\right)\cdot\left(3x-4\right)=0\\ \Leftrightarrow\left(x+3\right)\cdot\left(x-5+3x-4\right)=0\\ \Leftrightarrow\left(x+3\right)\cdot\left(4x-9\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+3=0\\4x-9=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\x=\frac{9}{4}\end{matrix}\right.\)
d)
\(\left(x+6\right)\cdot\left(3x-1\right)+x^2-36=0\\ \Leftrightarrow\left(x+6\right)\cdot\left(3x-1\right)+\left(x^2-36\right)=0\\ \Leftrightarrow\left(x+6\right)\cdot\left(3x-1\right)+\left(x+6\right)\cdot\left(x-6\right)=0\\ \Leftrightarrow\left(x+6\right)\cdot\left(3x-1+x-6\right)=0\\ \Leftrightarrow\left(x+6\right)\cdot\left(4x-7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+6=0\\4x-7=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-6\\x=\frac{7}{4}\end{matrix}\right.\)
e)
\(0.75x\cdot\left(x+5\right)=\left(x+5\right)\cdot\left(3-1.25x\right)\\ \Leftrightarrow0.75x\cdot\left(x+5\right)-\left(x+5\right)\cdot\left(3-1.25x\right)=0\\ \Leftrightarrow\left(x+5\right)\cdot\left(0.75x-3+1.25x\right)=0\\ \Leftrightarrow\left(x+5\right)\cdot\left(2x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+5=0\\2x-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-5\\x=\frac{3}{2}\end{matrix}\right.\)
a) ( 4x - 1 ) (x - 3) - ( x - 3 ) ( 5x + 2 ) = 0
<=> (x - 3)(4x - 1 - 5x - 2) = 0
<=> (x - 3)(-x - 3) = 0
<=> x = 3 hoặc x = -3
b) ( x + 3 ) ( x - 5 ) + ( x + 3 ) ( 3x - 4) = 0
<=> (x + 3)(x - 5 + 3x - 4) = 0
<=> (x + 3)(4x - 9) = 0
<=> x = -3 hoặc x = 9/4
c) ( x + 6 ) ( 3x - 1 )+ x2 - 36 = 0
<=> 3x^2 + 17x - 6 + x^2 - 36 = 0
<=> 4x^2 + 17x - 42 = 0
<=> 4x^2 + 24x - 7x - 42 = 0
<=> 4x(x + 6) - 7(x + 6) = 0
<=> (4x - 7)(x + 6) = 0
<=> x = -6 hoặc x = 7/4
d) ( x + 4 ) ( 5x + 9 ) - x2 + 16 = 0
<=> 5x^2 + 29x + 36 - x^2 + 16 = 0
<=> 4x^2 + 29x + 52 = 0
<=> 4x^2 + 16x + 13x + 42 = 0
<=> 4x(x + 4) + 13(x + 4) = 0
<=> (4x + 13)(x + 4) = 0
<=> x = -13/4 và x = -4
a/ (4x-1)(x-3)-(x-3)(5x+2)=0
<=> (x-3)(4x-1-5x-2)=0
<=> (x-3)(-x-3)=0
<=> x-3=0 hoặc -x-3=0
<=> x=3 hoặc x= -3
b/ (x+6)(3x-1)+ x^2 -36 =0
<=> (x+6)(3x-1) + (x-6)(x+6)=0
<=> (x+6)(3x-1+x-6)=0
<=> (x+6)(4x-7)=0
<=> x+6=o hoặc 4x-7=0
<=> x= -6 hoặc x= 7/4
c/ (x+3)(x+5)+(x+3)(3x-4)=0
<=> (x+3)(x+5+3x-4)=0
<=> (x+3)(4x+1)=0
<=> x+3=0 hoặc 4x+1=0
<=> x= -3 hoặc x=-1/4
c. x^2-5x +6 = 0
<=> x^2 - 5x = -6
<=> - 4x = -6
<=> x= -6/-4
Mình chỉ phân tích đa thức thành nhân tử thôi , phần còn lại bạn tự tính nha keo dài lắm
A) 2x2(x+3) - x(x+3) = 0 <=> x(x - 3)(2x-1)=0
B) (2x+5)2 - (x+2)2=0 <=> (x+3)(3x+7)=0
C) (x2-2x) - (3x-6)=0 <=> (x-2)(x-3)=0
D) (2x-7)(2x-7-6x+18)=0 <=> (2x-7)(-4x+11)=0
E) (x-2)(x+1) - (x-2)(x+2)=0 <=> (x-2)*(-1)=0 <=> x-2=0
G) (2x-3)(2x+2-5x)=0 <=> (2x-3)(-3x+2)=0
H) (1-x)(5x+3+3x-7)=0 <=> (1-x)(8x-4)=0
F) (x+6)*3x=0
I) (x-3)(4x-1-5x-2)=0 <=> (x-3)(-x-3)=0
K) (x+4)(5x+8)=0
H) (x+3)(4x-9)=0
c. x^2-5x+6=0
<=> x^2-5x=-6
<=> -4x=-6
<=> x=-6/-4
vậy tập nghiệm của pt là s={-6/-4}
a) (x + 6)(3x + 1) + x2 - 36 = 0
<=> 3x2 + x + 18x + 6 + x2 - 36 = 0
<=> 4x2 + 19x - 30 = 0
<=> 4x2 + 24x - 5x - 30 = 0
<=> 4x(x + 6) - 5(x + 6) = 0
<=> (x + 6)(4x - 5) = 0
<=> x + 6 = 0 hoặc 4x - 5 = 0
<=> x = -6 hoặc x = 5/4
Bài 1 mình đã làm xong rồi, anh em nào giúp mình bài 2 với!
Bài 2:
a: \(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
Bài 1 :
a, \(\left(4x-1\right)\left(x-3\right)-\left(x-3\right)\left(5x+2\right)=0\)
=> \(\left(x-3\right)\left(4x-1-5x-2\right)=0\)
=> \(\left(x-3\right)\left(-x-3\right)=0\)
=> \(\left[{}\begin{matrix}x-3=0\\-x-3=0\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
Vậy phương trình có nghiệm là \(x=\pm3\) .
b, \(\left(x+3\right)\left(x-5\right)+\left(x+3\right)\left(3x-4\right)=0\)
=> \(\left(x+3\right)\left(x-5+3x-4\right)=0\)
=> \(\left(x+3\right)\left(4x-9\right)=0\)
=> \(\left[{}\begin{matrix}x+3=0\\4x-9=0\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=-3\\x=\frac{9}{4}\end{matrix}\right.\)
Vậy phương trình có nghiệm là \(x=-3,x=\frac{9}{4}\) .
c, \(\left(x+6\right)\left(3x-1\right)+x^2-36=0\)
=> \(\left(x+6\right)\left(3x-1\right)+\left(x-6\right)\left(x+6\right)=0\)
=> \(\left(x+6\right)\left(3x-1+x-6\right)=0\)
=> \(\left(x+6\right)\left(4x-7\right)=0\)
=> \(\left[{}\begin{matrix}x+6=0\\4x-7=0\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=-6\\x=\frac{7}{4}\end{matrix}\right.\)
Vậy phương trình có nghiệm là \(x=-6,x=\frac{7}{4}\) .
a) ( 4x - 1 ) ( x - 3 ) - ( x - 3 ) ( 5x + 2 ) = 0
⇔ ( x - 3 ) ( 4x - 1 - 5x - 2 ) = 0
⇔ ( x - 3 ) ( -x - 3 ) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\-x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
Ý b) tương tự ý a) thôi.
c) ( x + 6 ) ( 3x - 1 ) + x2 - 36 = 0
⇔ ( x + 6 ) ( 3x - 1 ) + ( x + 6 ) ( x - 6 ) = 0
⇔ (x+6)(3x-1+x-6)=0
⇔ (x+6)(4x-7)=0
\(\Leftrightarrow\left[{}\begin{matrix}x+6=0\\4x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=\frac{7}{4}\end{matrix}\right.\)