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a, (3x - 2)(4x + 3) = (2 - 3x)(x - 1)
\(\Leftrightarrow\) (3x - 2)(4x + 3) - (2 - 3x)(x - 1) = 0
\(\Leftrightarrow\) (3x - 2)(4x + 3) + (3x - 2)(x - 1) = 0
\(\Leftrightarrow\) (3x - 2)(4x + 3 + x - 1) = 0
\(\Leftrightarrow\) (3x - 2)(5x + 2) = 0
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\5x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=\frac{-2}{5}\end{matrix}\right.\)
Vậy S = {\(\frac{2}{3}\); \(\frac{-2}{5}\)}
b, x2 + (x + 3)(5x - 7) = 9
\(\Leftrightarrow\) x2 - 9 + (x + 3)(5x - 7) = 0
\(\Leftrightarrow\) (x - 3)(x + 3) + (x + 3)(5x - 7) = 0
\(\Leftrightarrow\) (x + 3)(x - 3 + 5x - 7) = 0
\(\Leftrightarrow\) (x + 3)(6x - 10) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\6x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\frac{5}{3}\end{matrix}\right.\)
Vậy S = {-3; \(\frac{5}{3}\)}
c, 2x2 + 5x + 3 = 0
\(\Leftrightarrow\) 2x2 + 2x + 3x + 3 = 0
\(\Leftrightarrow\) 2x(x + 1) + 3(x + 1) = 0
\(\Leftrightarrow\) (x + 1)(2x + 3) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\frac{3}{2}\end{matrix}\right.\)
Vậy S = {-1; \(\frac{3}{2}\)}
d, \(\frac{3-2x}{2006}+\frac{3-2x}{2007}+\frac{3-2x}{2008}=\frac{3-2x}{2009}+\frac{3-2x}{2010}\)
\(\Leftrightarrow\) \(\frac{3-2x}{2006}+\frac{3-2x}{2007}+\frac{3-2x}{2008}-\frac{3-2x}{2009}-\frac{3-2x}{2010}=0\)
\(\Leftrightarrow\) (3 - 2x)\(\left(\frac{1}{2006}+\frac{1}{2007}+\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)\) = 0
\(\Leftrightarrow\) 3 - 2x = 0
\(\Leftrightarrow\) x = \(\frac{3}{2}\)
Vậy S = {\(\frac{3}{2}\)}
Chúc bn học tốt!!
\(5X\left(X-2020\right)+X=2020\)
\(\Leftrightarrow5X^2-10100X+X=2020\)
\(\Leftrightarrow5X^2-10099X=2020\)
\(\Leftrightarrow5X^2-10099X-2020=0\)
\(\Leftrightarrow5X^2-10100X+x-2020=0\)
\(\Leftrightarrow5X\left(X-2020\right)+X-2020=0\)
\(\Leftrightarrow\left(X-2020\right)\left(5X+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2020\\x=-\frac{1}{5}\end{cases}}\)
\(4\left(x-5\right)^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left[2\left(x-5\right)\right]^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left[2\left(x-5\right)-2x-1\right]\left[2\left(x-5\right)+2x+1\right]=0\)
\(\Leftrightarrow\left(2x-10-2x-1\right)\left(2x-10+2x+1\right)=0\)
\(\Leftrightarrow-11\left(4x-9\right)=0\)
\(\Leftrightarrow x=\frac{9}{4}\)
a) \(4\left(x-3\right)^2=9\left(2-3x\right)^2\)
\(\Leftrightarrow\left(2x-6\right)^2=\left(6-9x\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-6=6-9x\\2x-6=9x-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}11x=12\\7x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{12}{11}\\x=0\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là \(S=\left\{\frac{12}{11};0\right\}\)
b) \(ĐKXĐ:x\ne\pm1\)
\(\frac{x+1}{x-1}+\frac{x^2+3x-2}{1-x^2}=\frac{x-1}{x+1}\)
\(\Leftrightarrow\frac{x+1}{x-1}-\frac{x^2+3x-2}{x^2-1}-\frac{x-1}{x+1}=0\)
\(\Leftrightarrow\frac{\left(x+1\right)^2-x^2-3x+2-\left(x-1\right)^2}{x^2-1}=0\)
\(\Leftrightarrow\frac{x^2+2x+1-x^2-3x+2-x^2+2x-1}{x^2-1}=0\)
\(\Leftrightarrow-x^2+x+2=0\)
\(\Leftrightarrow x^2-x-2=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là \(S=\left\{2\right\}\)
Bạn đưa quá nhiều bài 1 lúc nên người ta giải được cũng chẳng ai muốn giải đâu, vì nhìn vào đã thấy ngộp rồi. Kinh nghiệm là muốn được giải quyết nhanh thì chỉ đăng 2-3 bài 1 lúc thôi
Bài 1:
a/ \(11-\left(2x+3\right)=3\left(x-4\right)\)
\(\Leftrightarrow11-2x-3=3x-12\)
\(\Leftrightarrow5x=20\)
\(\Rightarrow x=4\)
b/ \(5\left(2x-3\right)-4\left(5x-7\right)=19-2x\)
\(\Leftrightarrow10x-15-20x+28=19-2x\)
\(\Leftrightarrow8x=-6\)
\(\Rightarrow x=-\frac{3}{4}\)
c/
\(\frac{x}{3}-\frac{2x+1}{2}=\frac{x}{6}-x\)
\(\Leftrightarrow2x-3\left(2x+1\right)=x-6x\)
\(\Leftrightarrow x=3\)
d/
\(\frac{5x+2}{6}-\frac{8x-1}{3}=\frac{4x+2}{5}-5\)
\(\Leftrightarrow5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-150\)
\(\Leftrightarrow79x=158\)
\(\Rightarrow x=2\)
e/
\(\frac{2-6x}{5}-\frac{2+3x}{10}=7-\frac{6x+3}{4}\)
\(\Leftrightarrow4\left(2-6x\right)-2\left(2+3x\right)=140-5\left(6x+3\right)\)
\(\Leftrightarrow0=-121\) (vô lý)
Vậy pt vô nghiệm
f/
\(\frac{3x+2}{2}-\frac{3x+1}{6}=2x+\frac{5}{3}\)
\(\Leftrightarrow3\left(3x+2\right)-\left(3x+1\right)=12x+10\)
\(\Leftrightarrow6x=-5\)
\(\Rightarrow x=-\frac{5}{6}\)
Bài 1:
a) \(x^3-5x^2+8x-4\)
\(=x^3-4x^2+4x-x^2+4x-4\) \(=x\left(x^2-4x+4\right)-\left(x^2-4x+4\right)\)\(=\left(x-1\right)\left(x-2\right)^2\)
b) Ta có: \(\frac{A}{M}=\frac{10x^2-7x-5}{2x-3}=5x+4+\frac{7}{2x-3}\)
Với \(x\in Z\)thì \(A⋮M\)khi \(\frac{7}{2x-3}\in Z\)\(\Rightarrow7⋮\left(2x-3\right)\)\(\Rightarrow2x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
\(\Rightarrow=\left\{1;5;\pm2\right\}\)thì khi đó \(A⋮M\)
Các bài làm này có đúng ko ạ, ai đó duyệt giúp em, em cảm ơn.
Bài 1:
a)x3-5x2+8x-4=x3-4x2+4x-x2+4x-4
=x(x2-4x-4)-(x2-4x+4)
=(x-1) (x-2)2
b)Xét:
\(\frac{a}{b}-\frac{10x^2-7x-5}{2x-3}\)
=\(5x+4+\frac{7}{2x-3}\)
Với x thuộc Z thì A /\ B khi \(\frac{7}{2x-3}\) thuộc Z => 7 /\ (2x-3)
Mà Ư(7)={-1;1;-7;7} => x=5;-2;2;1 thì A /\ B
c)Biến đổi \(\frac{x}{y^3-1}-\frac{x}{x^3-1}=\frac{x^4-x-y^4+y}{\left(y^3-1\right)\left(x^3-1\right)}\)
=\(\frac{\left(x^4-y^4\right)\left(x-y\right)}{xy\left(y^2+y+1\right)\left(x^2+x+1\right)}\)(do x+y=1=>y-1=-x và x-1=-y)
=\(\frac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)-\left(x-y\right)}{xy\left[x^2y^2+y^2x+y^2+xy^2+xy+y+x^2+x+1\right]}\)
=\(\frac{\left(x-y\right)\left(x^2+y^2-1\right)}{xy\left[x^2y^2+xy\left(x+y\right)+x^2+y^2+xy+2\right]}\)
=\(\frac{\left(x-y\right)\left(x^2-x+y^2-y\right)}{xy\left[x^2y^2+\left(x+y\right)^2+2\right]}=\frac{\left(x-y\right)\left[x\left(x-1\right)+y\left(y-1\right)\right]}{xy\left(x^2y^2+3\right)}\)
=\(\frac{\left(x-y\right)\left[x\left(-y\right)+y\left(-x\right)\right]}{xy\left(x^2y^2+3\right)}=\frac{\left(x-y\right)\left(-2xy\right)}{xy\left(x^2y^2+3\right)}\)
=\(\frac{-2\left(x-y\right)}{x^2y^2+3}\)Suy ra điều phải chứng minh
Bài 2 )
a)(x2+x)2+4(x2+x)=12 đặt y=x2+x
y2+4y-12=0 <=>y2+6y-2y-12=0
<=>(y+6)(y-2)=0 <=> y=-6;y=2
>x2+x=-6 vô nghiệm vì x2+x+6 > 0 với mọi x
>x2+x=2 <=> x2+x-2=0 <=> x2+2x-x-2=0
<=>x(x+2)-(x+2)=0 <=>(x+2)(x-1) <=> x=-2;x-1
Vậy nghiệm của phương trình x=-2;x=1
b)\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}+\frac{x+4}{2005}+\frac{x+5}{2004}\)\(+\frac{x+6}{2003}\)
=\(\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)+\left(\frac{x+4}{2005}+1\right)\)\(+\left(\frac{x+5}{2004}+1\right)+\left(\frac{x+6}{2003}+1\right)\)
<=>\(\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\frac{x+2009}{2005}\)\(+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)
<=>\(\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}\)\(-\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x+2009}{2003}=0\)
Nhờ OLM xét giùm em vs ạ !
d, (x2 + 4x + 8)2 + 3x(x2 + 4x + 8) + 2x2 = 0
Đặt x2 + 4x + 8 = t ta được:
t2 + 3xt + 2x2 = 0
\(\Leftrightarrow\) t2 + xt + 2xt + 2x2 = 0
\(\Leftrightarrow\) t(t + x) + 2x(t + x) = 0
\(\Leftrightarrow\) (t + x)(t + 2x) = 0
Thay t = x2 + 4x + 8 ta được:
(x2 + 4x + 8 + x)(x2 + 4x + 8 + 2x) = 0
\(\Leftrightarrow\) (x2 + 5x + 8)[x(x + 4) + 2(x + 4)] = 0
\(\Leftrightarrow\) (x2 + 5x + \(\frac{25}{4}\) + \(\frac{7}{4}\))(x + 4)(x + 2) = 0
\(\Leftrightarrow\) [(x + \(\frac{5}{2}\))2 + \(\frac{7}{4}\)](x + 4)(x + 2) = 0
Vì (x + \(\frac{5}{2}\))2 + \(\frac{7}{4}\) > 0 với mọi x
\(\Rightarrow\left[{}\begin{matrix}x+4=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-2\end{matrix}\right.\)
Vậy S = {-4; -2}
Mình giúp bn phần khó thôi!
Chúc bn học tốt!!
c) \(\frac{1}{x-1}\)+\(\frac{2x^2-5}{x^3-1}\)=\(\frac{4}{x^2+x+1}\) (ĐKXĐ:x≠1)
⇔\(\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)+\(\frac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}\)=\(\frac{4\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
⇒x2+x+1+2x2-5=4x-4
⇔3x2-3x=0
⇔3x(x-1)=0
⇔x=0 (TMĐK) hoặc x=1 (loại)
Vậy tập nghiệm của phương trình đã cho là:S={0}
b) ( x2 - 9 ) . ( x - 7 ) = ( x + 3 ) . ( x2 + 6 )
<=> x3 - 7x2 - 9x + 63 = x3 + 6.x+ 3.x2 + 18
<=> x3 -7.x2 - 9.x + 63 - x3 + 6.x -3.x2 -18 =0
<=> -10.x2 - 15.x + 45 = 0
<=> 10.x2 + 15 .x - 45 = 0
<=> 5.( 2.x - 3 ) . ( x + 3 ) =0
<=> \(\orbr{\begin{cases}2.x-3=0\\x+3=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=\frac{3}{2}\\x=-3\end{cases}}\)
Vậy x = 3/2 ; -3
c) .....
a) \(\left(2x+1\right)\left(3x-2\right)=\left(2x+1\right)\left(5x-8\right)\)
\(\Leftrightarrow\)\(\left(2x+1\right)\left(3x-2\right)-\left(2x+1\right)\left(5x-8\right)=0\)
\(\Leftrightarrow\)\(\left(2x+1\right)\left(3x-2-5x+8\right)=0\)
\(\Leftrightarrow\)\(\left(2x+1\right)\left(6-2x\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}2x+1=0\\6-2x=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-0,5\\x=3\end{cases}}\)
Vậy...
b) \(ĐKXĐ:\) \(x\ne-2;\) \(x\ne4\)
\(\frac{3}{x+2}+\frac{2}{x-4}=0\)
\(\Leftrightarrow\)\(\frac{3\left(x-4\right)}{\left(x+2\right)\left(x-4\right)}+\frac{2\left(x+2\right)}{\left(x+2\right)\left(x-4\right)}=0\)
\(\Leftrightarrow\)\(\frac{3x-12+2x+4}{\left(x+2\right)\left(x-4\right)}=0\)
\(\Leftrightarrow\)\(\frac{5x-8}{\left(x+2\right)\left(x-4\right)}=0\)
\(\Rightarrow\)\(5x-8=0\)
\(\Leftrightarrow\)\(x=\frac{8}{5}\) (T/m đkxđ)
Vậy...
c) \(x^3+4x^2+4x+3=0\)
\(\Leftrightarrow\)\(x^3+3x^2+x^2+3x+x+3=0\)
\(\Leftrightarrow\)\(x^2\left(x+3\right)+x\left(x+3\right)+\left(x+3\right)=0\)
\(\Leftrightarrow\)\(\left(x+3\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\)\(x+3=0\) (do \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\) \(\forall x\))
\(\Leftrightarrow\)\(x=-3\)
Vậy...