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Ko viết lại đề
Câu 1: chia ra làm 3 trường hợp
Câu 2:
\(\left(x+2-x+2\right)\left(x+2\right)=0\)
\(4\left(x+2\right)=0\)
\(\Rightarrow x+2=0\)
\(x=-2\)
a)x2-20-x=0
<=>(x2-5x)+(4x-20)=0
<=>x(x-5)+4(x-5)=0
<=>(x-5)(x+4)=0
<=>x-5=0 hoặc x+4=0
<=>x=5 hoặc x=-4
b)(2x+3)2-(4x2-9)=0
<=>(2x+3)(2x+3)-(2x-3)(2x+3)=0
<=>(2x+3)(2x+3-2x+3)=0
<=>(2x+3).6=0
<=>2x+3=0
<=>2x=-3
<=>x=-1,5
c)(2x2+5x+3):(x+1)=4x-5
<=>2x2+5x+3=(4x-5)(x+1)
<=>2x2+5x+3=4x2-x-5
<=>4x2-x-5-2x2-5x-3=0
<=>2x2-6x-8=0
<=>x2-3x-4=0
<=>(x2-4x)+(x-4)=0
<=>x(x-4)+(x-4)=0
<=>(x-4)(x+1)=0
<=>x+1=0 hoặc x-4=0
<=>x=-1 hoặc x=4
a. \(2x\left(x+5\right)-x\left(3+2x\right)=26\Leftrightarrow2x^2+10x-3x-2x^2=26\Leftrightarrow7x=26\Leftrightarrow x=\dfrac{26}{7}\)
Vậy \(x=\dfrac{26}{7}\)
b. \(5x\left(x-1\right)=x-1\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x-1=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\5x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
c. \(2\left(x+5\right)-x^2-5x=0\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\Leftrightarrow\left(x+5\right)\left(2-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
d. \(\left(2x-3\right)^2-\left(x+5\right)^2=0\Leftrightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\Leftrightarrow\left(x-8\right)\left(3x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x-8=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\3x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)
e. \(3x^3-48x=0\Leftrightarrow3x\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}3x=0\\x^2-16=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm4\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=0\\x=\pm4\end{matrix}\right.\)
f. \(x^3+x^2-4x=4\Leftrightarrow x^3+x^2-4x-4=0\Leftrightarrow\left(x^2-4x+4\right)+\left(x^3-8\right)=0\Leftrightarrow\left(x-2\right)^2+\left(x-2\right)\left(x^2+2x+4\right)=0\Leftrightarrow\left(x-2\right)\left(x-2+x^2+2x+4\right)=0\left(x-2\right)\left(x^2+3x+2\right)=0\Leftrightarrow\left(x-2\right)\left(x^2+x+2x+2\right)=0\Leftrightarrow\left(x-2\right)\left[x\left(x+1\right)+2\left(x+1\right)\right]=0\Leftrightarrow\left(x-2\right)\left(x+1\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\\x=-2\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-1\\x=\pm2\end{matrix}\right.\)
g. \(\left(x-1\right)\left(2x+3\right)-x\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(2x+3-x\right)=0\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
h. \(x^2-4x+8=2x-1\Leftrightarrow x^2-4x+8-2x+1=0\Leftrightarrow x^2-6x+9=0\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy \(x=3\)
__________________________Chúc bạn học tốt____________________________
Bài làm :
a) x( 2x - 7 ) - 4x + 14 = 0
<=> x( 2x - 7 ) - 2( 2x - 7 ) = 0
<=> ( 2x - 7 )( x - 2 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}2x-7=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=2\end{cases}}\)
b) Sửa đề : 5x3 + x2 - 4x + 9 = 0
<=>( 5x3 + 5 ) + (x2 - 4x +4)=0
<=> 5(x3 + 1) + (x-2)2 = 0
<=> 5(x+1)(x2 - x +1) + (x+2)2 =0
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)
c) 3x3 - 7x2 + 6x - 14 = 0
<=> 3x2( x - 7/3 ) + 6( x - 7/3 ) = 0
<=> ( x - 7/3 )( 3x2 + 6 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{7}{3}=0\\3x^2+6=0\end{cases}}\Leftrightarrow x=\frac{7}{3}\)
d) 5x2 - 5x = 3( x - 1 )
<=> 5x( x - 1 ) - 3( x - 1 ) = 0
<=> ( x - 1 )( 5x - 3 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\5x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{3}{5}\end{cases}}\)
e) 4x2 - 25 - ( 4x - 10 ) = 0
<=> ( 2x - 5 )( 2x + 5 ) - 2( 2x - 5 ) = 0
<=> ( 2x - 5 )( 2x + 5 - 2 ) = 0
<=> ( 2x - 5 )( 2x + 3 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}2x-5=0\\2x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{3}{2}\end{cases}}\)
f) x3 + 27 + ( x + 3 )( x - 9 ) = 0
<=> ( x + 3 )( x2 - 3x + 9 ) + ( x + 3 )( x - 9 ) = 0
<=> ( x + 3 )( x2 - 3x + 9 + x - 9 ) = 0
<=> ( x + 3 )( x2 - 2x ) = 0
<=> x( x + 3 )( x - 2 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}\\\end{cases}}\begin{cases}x=0\\x=-3\\x=2\end{cases}\)
a) x( 2x - 7 ) - 4x + 14 = 0
<=> x( 2x - 7 ) - 2( 2x - 7 ) = 0
<=> ( 2x - 7 )( x - 2 ) = 0
<=> \(\orbr{\begin{cases}2x-7=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=2\end{cases}}\)
b) 5x3 + x2 - 4x - 9 = 0 ( đề sai )
c) 3x3 - 7x2 + 6x - 14 = 0
<=> 3x2( x - 7/3 ) + 6( x - 7/3 ) = 0
<=> ( x - 7/3 )( 3x2 + 6 ) = 0
<=> \(\orbr{\begin{cases}x-\frac{7}{3}=0\\3x^2+6=0\end{cases}}\Leftrightarrow x=\frac{7}{3}\)( do 3x2 + 6 ≥ 6 > 0 với mọi x )
d) 5x2 - 5x = 3( x - 1 )
<=> 5x( x - 1 ) - 3( x - 1 ) = 0
<=> ( x - 1 )( 5x - 3 ) = 0
<=> \(\orbr{\begin{cases}x-1=0\\5x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{3}{5}\end{cases}}\)
e) 4x2 - 25 - ( 4x - 10 ) = 0
<=> ( 2x - 5 )( 2x + 5 ) - 2( 2x - 5 ) = 0
<=> ( 2x - 5 )( 2x + 5 - 2 ) = 0
<=> ( 2x - 5 )( 2x + 3 ) = 0
<=> \(\orbr{\begin{cases}2x-5=0\\2x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{3}{2}\end{cases}}\)
f) x3 + 27 + ( x + 3 )( x - 9 ) = 0
<=> ( x + 3 )( x2 - 3x + 9 ) + ( x + 3 )( x - 9 ) = 0
<=> ( x + 3 )( x2 - 3x + 9 + x - 9 ) = 0
<=> ( x + 3 )( x2 - 2x ) = 0
<=> x( x + 3 )( x - 2 ) = 0
<=> x = 0 hoặc x + 3 = 0 hoặc x - 2 = 0
<=> x = 0 hoặc x = -3 hoặc x = 2
Ví dụ cho bạn một bài, còn lại tương tự.
a)Ta có: \(3x^4-5x^3+8x^2-5x+3\)
\(=3x^2\left(x-\frac{5}{6}\right)^2+\frac{71}{12}\left(x-\frac{30}{71}\right)^2+\frac{138}{71}>0\)
Vậy phương trình vô nghiệm.
1)⇔x2+1x-3x+3=0
⇔x(x+1)-3(x+1)=0
⇔(x+1)(x-3)=0
⇔x+1=0 hoặc x-3=0
⇔x=-1 hoặc x=3
4)⇔x(1+5x)=0
⇔x=0 hoặc 1+5x=0
⇔x=0 hoặc 5x=-1
⇔x=0 hoặc x=-0.2
a) 2x3-50x=2x(x2-25)=0
=> \(\left\{{}\begin{matrix}2x=0=>x=0\\x^2-25=0=>x=\pm5\end{matrix}\right.\)
b) x3-8=(x-2)3
<=>(x-2)(x2+2x+4)=(x-2)(x2-4x+4)
<=>(x-2)(x2+2x+4)-(x-2)(x2-4x+4)=0
<=>(x-2)(x2-x2+2x+4x+4-4)
<=>6x(x-2)=0
\(=>\left\{{}\begin{matrix}6x=0=>x=0\\x-2=0=>x=2\end{matrix}\right.\)
c) x3+5x2-4x-20=0
<=>x2(x+5)-4(x+5)=0
<=>(x2-4)(x+5)=0
\(=>\left\{{}\begin{matrix}x^2-4=0=>x=\pm2\\x+5=0=>x=-5\end{matrix}\right.\)