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8 tháng 8 2018

1) \(-6x^2-x+7=0\)

\(\Leftrightarrow-6x^2+6x-7x+7=0\)

\(\Leftrightarrow\left(-6x^2+6x\right)-\left(7x-7\right)=0\)

\(\Leftrightarrow-6x\left(x-1\right)-7\left(x-1\right)=0\)

\(\Leftrightarrow\left(-6x-7\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-6x-7=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{6}\\x=1\end{matrix}\right.\)

2) \(-4x^2-5x+9=0\)

\(\Leftrightarrow-4x^2+4x-9x+9=0\)

\(\Leftrightarrow\left(-4x^2+4x\right)-\left(9x-9\right)=0\)

\(\Leftrightarrow-4x\left(x-1\right)-9\left(x-1\right)=0\)

\(\Leftrightarrow\left(-4x-9\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-4x-9=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{9}{4}\\x=1\end{matrix}\right.\)

3) \(x^2+3x-4=0\)

\(\Leftrightarrow x^2-x+4x-4=0\)

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

\(\Leftrightarrow x\left(x-1\right)+4\left(x-1\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=1\end{matrix}\right.\)

4) \(x^2-6x-7=0\)

\(\Leftrightarrow x^2+x-7x-7=0\)

\(\Leftrightarrow\left(x^2+x\right)-\left(7x+7\right)=0\)

\(\Leftrightarrow x\left(x+1\right)-7\left(x+1\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)

5) \(x^2+5x+4=0\)

\(\Leftrightarrow x^2+x+4x+4=0\)

\(\Leftrightarrow\left(x^2+x\right)+\left(4x+4\right)=0\)

\(\Leftrightarrow x\left(x+1\right)+4\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)

a) \(-6x^2-x+7=0\)

\(\Leftrightarrow-6x^2+6x-7x+7=0\)

\(\Leftrightarrow-6x\left(x-1\right)-7\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(-6x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-7}{6}\end{matrix}\right.\)

b) \(-4x^2-5x+9=0\)

\(\Leftrightarrow-4x^2+4x-9x+9=0\)

\(\Leftrightarrow-4x\left(x-1\right)-9\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(-4x-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2,25\end{matrix}\right.\)

c) \(x^2+3x-4=0\)

\(\Leftrightarrow x^2-x+4x-4=0\)

\(\Leftrightarrow x\left(x-1\right)+4\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)

d) \(x^2-6x-7=0\)

\(\Leftrightarrow x^2+x-7x-7=0\)

\(\Leftrightarrow x\left(x+1\right)-7\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=7\end{matrix}\right.\)

e) \(x^2+5x+4=0\)

\(\Leftrightarrow x^2+x+4x+4=0\)

\(\Leftrightarrow x\left(x+1\right)+4\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-4\end{matrix}\right.\)

1: \(\Leftrightarrow3x+4x=4\)

=>7x=4

hay x=4/7

2: \(\Leftrightarrow3x-5x-5^3:5^2=0\)

=>-2x=5

=>x=-5/2

26 tháng 7 2018

1,\(x^4-x=0\\ ->x\left(x-1\right)\left(x^2+x+1\right)=0\\ ->\left(......\right)\)

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

3,\(x^5+x^2\\ ->x^2\left(x^3+1\right)\\ ->x^2\left(x+1\right)\left(x^2-x+1\right)\\ ->.......\)

4\(3x\left(x-20\right)-x+20=0->\left(3x-1\right)\left(x-20\right)=0->.....\)

8 tháng 8 2018

1) \(x^2-8x+7=0\)

\(\Leftrightarrow x^2-7x-x+7=0\)

\(\Leftrightarrow\left(x^2-7x\right)-\left(x-7\right)=0\)

\(\Leftrightarrow x\left(x-7\right)-\left(x-7\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=7\end{matrix}\right.\)

2) \(5x^2-11x+6=0\)

\(\Leftrightarrow5x^2-5x-6x+6=0\)

\(\Leftrightarrow\left(5x^2-5x\right)-\left(6x-6\right)=0\)

\(\Leftrightarrow5x\left(x-1\right)-6\left(x-1\right)=0\)

\(\Leftrightarrow\left(5x-6\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-6=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\\x=1\end{matrix}\right.\)

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

\(\Leftrightarrow2x^2-2x-x+1=0\)

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

\(\Leftrightarrow2x\left(x-1\right)-\left(x-1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)

4) \(x^2+7x-8=0\)

\(\Leftrightarrow x^2+8x-x-8=0\)

\(\Leftrightarrow\left(x^2+8x\right)-\left(x+8\right)=0\)

\(\Leftrightarrow x\left(x+8\right)-\left(x+8\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-8\end{matrix}\right.\)

5) \(3x^2+7x-10=0\)

\(\Leftrightarrow3x^2-3x+10x-10=0\)

\(\Leftrightarrow\left(3x^2-3x\right)+\left(10x-10\right)=0\)

\(\Leftrightarrow3x\left(x-1\right)+10\left(x-1\right)=0\)

\(\Leftrightarrow\left(3x+10\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+10=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{10}{3}\\x=1\end{matrix}\right.\)

8 tháng 8 2018

1) \(5x^2-2x-7=0\)

\(\Leftrightarrow5x^2+5x-7x-7=0\)

\(\Leftrightarrow5x\left(x+1\right)-7\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(5x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\5x-7=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{7}{5}\end{matrix}\right.\)

2) \(x^2-10x-11=0\)

\(\Leftrightarrow x^2+x-11x-11=0\)

\(\Leftrightarrow x\left(x+1\right)-11\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-11=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=11\end{matrix}\right.\)

3) \(x^2-7x-8=0\)

\(\Leftrightarrow x^2+x-8x-8=0\)

\(\Leftrightarrow x\left(x+1\right)-8\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-8=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=8\end{matrix}\right.\)

4) \(x^2+9x+8=0\)

\(\Leftrightarrow x^2+x+8x+8=0\)

\(\Leftrightarrow x\left(x+1\right)+8\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+8=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-8\end{matrix}\right.\)

5) \(x^2-5x-6=0\)

\(\Leftrightarrow x^2+x-6x-6=0\)

\(\Leftrightarrow x\left(x+1\right)-6\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=6\end{matrix}\right.\)

19 tháng 7 2016

a)      \(2\left(x+5\right)-x^2-5x=0\)

  \(\Leftrightarrow2x+10-x^2-5x=0\)

 \(\Leftrightarrow-x^2-3x+10=0\)

\(\Leftrightarrow x^2+3x-10=0\)

 \(\Leftrightarrow x^2-2x+5x-10=0\)

\(\Leftrightarrow x\left(x-2\right)+5\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}}\)

b) \(x^3-6x^2+12x-8=0\)

\(\Leftrightarrow\left(x^3-8\right)-\left(6x^2-12x\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)-6x\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4-6x\right)=0\)

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

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

\(\Leftrightarrow\left(x-2\right)^3=0\)

\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)

c)\(16x^2-9\left(x+1\right)^2=0\)

\(\Leftrightarrow\left(4x\right)^2-\left[3\left(x+1\right)\right]^2=0\)

\(\Leftrightarrow\left(4x-3x-1\right)\left(4x+3x+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(7x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\7x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{7}\end{cases}}}\)

d) \(x^3+x=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)

e)\(x^2-2x-3=0\)

\(\Leftrightarrow x^2+x-3x-3=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}}\)

19 tháng 7 2016

Cảm ơn bạn nha

4 tháng 12 2018

1) \(16x\left(2-x\right)-\left(4x-5\right)^2=0\)

\(32x-16x^2-16x^2+40x-25=0\)

\(72x-16x^2-25=0\)

Đề sai ko bạn nhonhung

2) \(\left(x-7\right)^2+3=\left(x-2\right)\left(x+2\right)\)

\(\left(x^2-14x+7\right)+3-\left(x-2\right)\left(x+2\right)=0\)

\(x^2-14x+7+3-x^2+4=0\)

\(-14x+14=0\)

\(x=1\)

3) \(\left(2x-3\right)^2-\left(7x-2x\right)^2=2\)

\(\left(2x-3\right)^2-\left(5x\right)^2=2\)

\(\left(2x-3-5x\right)\left(2x-3+5x\right)=2\)

\(\left(-3x-3\right)\left(7x-3\right)=2\)

=> lập bảng tìm x

4) \(\left(5x-7\right)^2-\left(1-3x\right)^2=16x\left(x-3\right)\)

\(25x^2-70x+49-9x^2+6x-1-16x^2+48x=0\)

\(-16x+48=0\)

\(x=3\)

4 tháng 12 2018

đề có vẻ sai nhiều quá :((

11 tháng 2 2018

a, (3x+1)(7x+3)=(5x-7)(3x+1)

<=> (3x+1)(7x+3)-(5x-7)(3x+1)=0

<=> (3x+1)(7x+3-5x+7)=0

<=> (3x+1)(2x+10)=0

<=> 2(3x+1)(x+5)=0

=> 3x+1=0 hoặc x+5=0

=> x= -1/3 hoặc x=-5

Vậy...

27 tháng 5 2018

a) (3x - 2)(4x + 5) = 0

⇔ 3x - 2 = 0 hoặc 4x + 5 = 0

1) 3x - 2 = 0 ⇔ 3x = 2 ⇔ x = 2/3

2) 4x + 5 = 0 ⇔ 4x = -5 ⇔ x = -5/4

Vậy phương trình có tập nghiệm S = {2/3;−5/4}

b) (2,3x - 6,9)(0,1x + 2) = 0

⇔ 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0

1) 2,3x - 6,9 = 0 ⇔ 2,3x = 6,9 ⇔ x = 3

2) 0,1x + 2 = 0 ⇔ 0,1x = -2 ⇔ x = -20.

Vậy phương trình có tập hợp nghiệm S = {3;-20}

c) (4x + 2)(x2 +  1) = 0 ⇔ 4x + 2 = 0 hoặc x2 +  1 = 0

1) 4x + 2 = 0 ⇔ 4x = -2 ⇔ x = −1/2

2) x2 +  1 = 0 ⇔ x2 = -1 (vô lí vì x2 ≥ 0)

Vậy phương trình có tập hợp nghiệm S = {−1/2}

d) (2x + 7)(x - 5)(5x + 1) = 0

⇔ 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0

1) 2x + 7 = 0 ⇔ 2x = -7 ⇔ x = −7/2

2) x - 5 = 0 ⇔ x = 5

3) 5x + 1 = 0 ⇔ 5x = -1 ⇔ x = −1/5

Vậy phương trình có tập nghiệm là S = {−7/2;5;−1/5}