tính a) (2x-3).(x+1)-2x(2-x)-4x^2+5x
b) x^3-6x^2+9x+14:x-7
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\(a,6x-4=5x\\ \Leftrightarrow x-4=0\\ \Leftrightarrow x=4\\ b,\dfrac{2x+3}{3}=\dfrac{5-4x}{2}\\ \Leftrightarrow2\left(2x+3\right)=3\left(5-4x\right)\\ \Leftrightarrow4x+6=15-12x\\ \Leftrightarrow16x-9=0\\ \Leftrightarrow x=\dfrac{9}{16}\\ c,\left(x+7\right)\left(x-10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+7=0\\x-10=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-7\\x=10\end{matrix}\right.\)
d, ĐKXĐ:\(x\ne\pm3\)
\(\dfrac{2}{x-3}+\dfrac{3}{x+3}=\dfrac{3x+5}{x^2-9}\\ \Leftrightarrow\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{3\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{3x+5}{\left(x+3\right)\left(x-3\right)}=0\\ \Leftrightarrow\dfrac{2x+6+3x-9-3x-5}{\left(x+3\right)\left(x-3\right)}=0\\ \Rightarrow2x-8=0\\ \Leftrightarrow x=4\left(tm\right)\)
a.6x-4=5x <=> x=4
b.\(\dfrac{2x+3}{3}=\dfrac{5-4x}{2}\)
\(\Leftrightarrow\dfrac{2\left(2x+3\right)}{6}=\dfrac{3\left(5-4x\right)}{6}\)
\(\Leftrightarrow2\left(2x+3\right)=3\left(5-4x\right)\)
\(\Leftrightarrow4x+6=15-12x\)
\(\Leftrightarrow16x=11\)
\(\Leftrightarrow x=\dfrac{11}{16}\)
c.(x+7)(x-10)=0
\(\Leftrightarrow\left[{}\begin{matrix}x+7=0\\x-10=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=10\end{matrix}\right.\)
d.\(ĐK:x\ne\pm3\)
\(\Rightarrow\dfrac{2}{x-3}+\dfrac{3}{x+3}=\dfrac{3x+5}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{2\left(x+3\right)+3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x+5}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow2\left(x+3\right)+3\left(x-3\right)=3x+5\)
\(\Leftrightarrow2x+6+3x-9-3x-5=0\)
\(\Leftrightarrow2x-8=0\)
\(\Leftrightarrow2x=8\)
\(\Leftrightarrow x=4\left(tm\right)\)
a/ (2x + 1)(4x – 3) – 6x(x + 5) – 2x(x – 7) + 18x
=8x^2-6x+4x-3-6x^2-30x-2x^2+14x+18x
=-3
vậy...
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x2 - 16x - 34 = 10x2 + 3x - 34
=> 10x2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0
hoặc 10x - 19 = 0 => 10x = 19 => x = 19/10
Vậy x = 0 ; x = 19/10
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x 2 - 16x - 34 = 10x 2 + 3x - 34
=> 10x 2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0 hoặc 10x - 19 = 0
=> 10x = 19
=> x = 19/10
Vậy x = 0 ; x = 19/10
a) giải pt ra ta được : x=-1
b) giải pt ra ta được : x=2
c)giải pt ra ta được : x vô ngiệm
d)giải pt ra ta được : x=vô ngiệm
~~~~~~~~~~~ai đi ngang qua nhớ để lại k ~~~~~~~~~~~~~
~~~~~~~~~~~~ Chúc bạn sớm kiếm được nhiều điểm hỏi đáp ~~~~~~~~~~~~~~~~~~~
Tim x,
a,2x^4-6x^3+x^2+6x-3=0
b,x^3-9x^2+26x+24=0
c, P= 2x^4 - 4x^3 + 6x^2 - 4x + 5 biet rang x^2 - x=7
a)\(2x^4-6x^3+x^2+6x-3=0\)
\(\Leftrightarrow2x^4-6x^3+3x^2-2x^2+6x-3=0\)
\(\Leftrightarrow x^2\left(2x^2-6x+3\right)-\left(2x^2-6x+3\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(2x^2-6x+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(2x^2-6x+3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x+1=0\\2x^2-6x+3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-1\\\Delta_{2x^2-6x+3}=\left(-6\right)^2-4\left(2.3\right)=12\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-1\\x_{1,2}=\frac{6\pm\sqrt{12}}{4}\end{array}\right.\)
b)\(x^3+9x^2+26x+24=0\)
\(\Leftrightarrow x^3+5x^2+6x+4x^2+20x+24=0\)
\(\Leftrightarrow x\left(x^2+5x+6\right)+4\left(x^2+5x+6\right)=0\)
\(\Leftrightarrow\left(x^2+5x+6\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x+2=0\\x+3=0\\x+4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-2\\x=-3\\x=-4\end{array}\right.\)
a) \(\sqrt[]{x^2-4x+4}=x+3\)
\(\Leftrightarrow\sqrt[]{\left(x-2\right)^2}=x+3\)
\(\Leftrightarrow\left|x-2\right|=x+3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=x+3\\x-2=-\left(x+3\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}0x=5\left(loại\right)\\x-2=-x-3\end{matrix}\right.\)
\(\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)
b) \(2x^2-\sqrt[]{9x^2-6x+1}=5\)
\(\Leftrightarrow2x^2-\sqrt[]{\left(3x-1\right)^2}=5\)
\(\Leftrightarrow2x^2-\left|3x-1\right|=5\)
\(\Leftrightarrow\left|3x-1\right|=2x^2-5\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=2x^2-5\\3x-1=-2x^2+5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x^2-3x-4=0\left(1\right)\\2x^2+3x-6=0\left(2\right)\end{matrix}\right.\)
Giải pt (1)
\(\Delta=9+32=41>0\)
Pt \(\left(1\right)\) \(\Leftrightarrow x=\dfrac{3\pm\sqrt[]{41}}{4}\)
Giải pt (2)
\(\Delta=9+48=57>0\)
Pt \(\left(2\right)\) \(\Leftrightarrow x=\dfrac{-3\pm\sqrt[]{57}}{4}\)
Vậy nghiệm pt là \(\left[{}\begin{matrix}x=\dfrac{3\pm\sqrt[]{41}}{4}\\x=\dfrac{-3\pm\sqrt[]{57}}{4}\end{matrix}\right.\)
Bài làm
a) ( 2x - 3 ).( x + 1 ) - 2x( 2 - x ) - 4x2 + 5x
= 2x2 + 2x - 3x - 3 - 4x + 2x2 - 4x2 + 5x
= -3
b) x3 - 6x2 + 9x + 14 : ( x - 7 )
Đặt cột chia ta được:
= x2 + x + 16 ( dư 126 )
Nhưng nếu đề thế này tính dễ hơn.
x3 - 6x2 - 9x + 14 : ( x - 7 )
= x3 - 8x2 + 7x + 2x2 - 16x + 14 : ( x - 7 )
= ( 7x + 14 ) + ( x3 - 8x2 + 2x2 - 16x ) : ( x - 7 )
= 7( x + 2 ) + x( x2 - 8x + 2x - 16 ) : ( x - 7 )
= 7( x + 2 ) + x[ ( x2 - 8x ) + ( 2x - 16 ) ] : ( x - 7 )
= 7( x + 2 ) + x[ x( x - 8 ) + 2( x - 8 ) : ( x - 7 )
= 7( x + 2 ) + x( x + 2 )( x - 8 ) : ( x - 7 )
= ( x + 2 )[ 7 + x( x - 8 )] : ( x - 7 )
= ( x + 2 )( x2 - 8x + 7 ) : ( x - 7 )
= ( x + 2 )( x2 - 7x - x + 7 ) : ( x - 7 )
= ( x + 2 )[ ( x2 - 7x ) - ( x - 7 ) ] : ( x - 7 )
= ( x + 2 )[ x( x - 7 ) - ( x - 7 ) ] : ( x - 7 )
= ( x - 2 )( x - 1 )( x - 7 ) : ( x - 7 )
= ( x - 2 )( x - 1 )
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