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5 tháng 8 2016
ap dung cong thuc: a/b = c/d <=> ad= bc <=> c = ad/b
A = (4x2-7x+3)(x2+2x+1)/(x2-1)
11 tháng 7 2017
\(\left(\frac{9}{x.x^2-9.x}+\frac{1}{x+_{ }3}\right):\left(\frac{x-3}{x.3+x^2}-\frac{x}{3.x+9}\right)\) đk (x\(\ne\)o; công trừ 3)
<=>\(9+\frac{x.\left(x-3\right)}{x.\left(x^2-9\right)}\):\(\frac{3.\left(x-3\right)-x^2}{3x.\left(x+3\right)}\)
<=>\(-\frac{3}{x-3}=\frac{3}{3-x}\)
TA
12 tháng 7 2017
Bạn ơi mk k hiểu sao lại ra bước 2 ... bạn giải chi tiết giùm mk nha
dù sao cx cảm ơn bạn đã giúp mk
LA
21 tháng 6 2017
b)(2x - 1)^2 - (2x + 5) (2x - 5 ) = 18
4x 2 -4x+1-4x 2+25=18
26-4x=18
4x=8
x=2
21 tháng 6 2017
a,27x-18=2x-3x^2
<=> 3x^2-2x+27-18x=0
<=> 3x^2-20x+27=0
\(\Delta\)= 20^2-4-12.27
tính \(\Delta\)rồi tìm x1 ,x2
11 giờ trước (12:04)
Bài 4:
a: \(2x^4+18x^2=0\)
=>\(2x^2\left(x^2+9\right)=0\)
=>\(x^2=0\) (Vì \(2\left(x^2+9\right)=2x^2+18\ge18>0\forall x\) )
=>x=0
b: (x-5)(x+5)-15x+75=0
=>(x-5)(x+5)-15(x-5)=0
=>(x-5)(x+5-15)=0
=>(x-5)(x-10)=0
=>\(\left[\begin{array}{l}x-5=0\\ x-10=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5\\ x=10\end{array}\right.\)
c: \(x^4=x^2\)
=>\(x^4-x^2=0\)
=>\(x^2\left(x^2-1\right)=0\)
=>\(\left[\begin{array}{l}x^2=0\\ x^2-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x^2=0\\ x^2=1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=1\\ x=-1\end{array}\right.\)
d: \(12x\left(6x-1\right)-24x^2=0\)
=>12x(6x-1-2x)=0
=>x(4x-1)=0
=>\(\left[\begin{array}{l}x=0\\ 4x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=\frac14\end{array}\right.\)
Bài 2:
a: 4x-16+3y(4-x)
=4(x-4)-3y(x-4)
=(x-4)(4-3y)
b: \(9y^2-6y+1=\left(3y\right)^2-2\cdot3y\cdot1+1^2=\left(3y-1\right)^2\)
c: \(25x^2-4=\left(5x\right)^2-2^2=\left(5x-2\right)\left(5x+2\right)\)
d: \(x^2-12x+36=x^2-2\cdot x\cdot6+6^2=\left(x-6\right)^2\)
e: \(8x^3+36x^2+54x+27\)
\(=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot3+3\cdot2x\cdot3^2+3^3\)
\(=\left(2x+3\right)^3\)
f: \(\left(2x-5\right)^2-\left(2x+y\right)^2\)
=(2x-5-2x-y)(2x-5+2x+y)
=(-y-5)(4x+y-5)
g: \(\left(2x-y\right)^3+\left(2x+y\right)^3\)
\(=8x^3-12x^2y+6xy^2-y^3+8x^3+12x^2y+6xy^2+y^3\)
\(=16x^3+12xy^2=4x\left(4x^2+3y^2\right)\)
Câu 1:
a: \(6x^2-72x=0\)
=>\(6\left(x^2-12x\right)=0\)
=>\(x^2-12x=0\)
=>x(x-12)=0
=>\(\left[\begin{array}{l}x=0\\ x-12=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=12\end{array}\right.\)
b: \(-2x^4+16x=0\)
=>\(-2x\left(x^3-8\right)=0\)
=>\(x\left(x^3-8\right)=0\)
=>\(\left[\begin{array}{l}x=0\\ x^3-8=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x^3=8\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=2\end{array}\right.\)
c: \(\left(2x-1\right)^3-8x\left(x-3\right)\cdot\left(x+3\right)=-1\)
=>\(8x^3-12x^2+6x-1-8x\cdot\left(x^2-9\right)=-1\)
=>\(8x^3-12x^2+6x-1-8x^3+72x=-1\)
=>\(-12x^2+78x=0\)
=>-6x(2x-13)=0
=>x(2x-13)=0
=>\(\left[\begin{array}{l}x=0\\ 2x-13=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=\frac{13}{2}\end{array}\right.\)
d: \(x\left(x-5\right)-\left(x-3\right)^2=0\)
=>\(x^2-5x-\left(x^2-6x+9\right)=0\)
=>\(x^2-5x-x^2+6x-9=0\)
=>x-9=0
=>x=9
e: \(x\left(x-5\right)+3\left(x-5\right)=0\)
=>(x-5)(x+3)=0
=>\(\left[\begin{array}{l}x-5=0\\ x+3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5\\ x=-3\end{array}\right.\)
f: 2x(x-8)-5(8-x)=0
=>2x(x-8)+5(x-8)=0
=>(x-8)(2x+5)=0
=>\(\left[\begin{array}{l}x-8=0\\ 2x+5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=8\\ x=-\frac52\end{array}\right.\)
g: \(30x-15x^2=0\)
=>15x(2-x)=0
=>x(2-x)=0
=>\(\left[\begin{array}{l}x=0\\ 2-x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=2\end{array}\right.\)
h: \(-4x^3-12x=0\)
=>\(-4x\left(x^2+3\right)=0\)
=>x=0
DD
Đoàn Đức Hà
Giáo viên
8 tháng 5 2021
\(\frac{x+2}{2019}+\frac{x+3}{2018}=\frac{x+4}{2017}+\frac{x}{2021}\)
\(\Leftrightarrow\frac{x+2}{2019}+1+\frac{x+3}{2018}+1=\frac{x+4}{2017}+1+\frac{x}{2021}+1\)
\(\Leftrightarrow\frac{x+2021}{2019}+\frac{x+2021}{2018}=\frac{x+2021}{2017}+\frac{x+2021}{2021}\)
\(\Leftrightarrow x+2021=0\)
\(\Leftrightarrow x=-2021\)
25 tháng 7
x+2+2018x+3=2017x+4+2021x
\(\Leftrightarrow \frac{x + 2}{2019} + 1 + \frac{x + 3}{2018} + 1 = \frac{x + 4}{2017} + 1 + \frac{x}{2021} + 1\)
\(\Leftrightarrow \frac{x + 2021}{2019} + \frac{x + 2021}{2018} = \frac{x + 2021}{2017} + \frac{x + 2021}{2021}\)
\(\Leftrightarrow x + 2021 = 0\)
\(\Leftrightarrow x = - 2021\)
TM
17 tháng 7 2017
a)\(B=\left(\frac{x-2}{x^2+2x}+\frac{1}{x+2}\right).\frac{x+1}{x-1}=\left(\frac{x^2-2}{x^2+2x}+\frac{x}{x^2+2x}\right).\frac{x+1}{x-1}=\frac{x^2+x-2}{x^2+2x}.\frac{x+1}{x-1}\)
\(=\frac{x^2-x+2x-2}{x\left(x+2\right)}.\frac{x+1}{x-1}=\frac{x\left(x-1\right)+2\left(x-1\right)}{x\left(x+2\right)}.\frac{x+1}{x-1}=\frac{\left(x-1\right)\left(x+2\right)}{x\left(x+2\right)}.\frac{x+1}{x-1}=\frac{x+1}{x}\)
b)\(2B=2x+5\Leftrightarrow2.\frac{x+1}{x}=2x+5\Leftrightarrow\frac{2x+2}{x}=2x+5\Leftrightarrow2x+2=2x^2+5x\)
\(\Leftrightarrow0=2x^2+3x-2\Leftrightarrow2x^2+4x-x-2=0\Leftrightarrow2x\left(x+2\right)-\left(x+2\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x+2\right)=0\Leftrightarrow\orbr{\begin{cases}2x-1=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-2\end{cases}}\)
KM
1
31 tháng 8 2018
\(-2x^2+5x=16\)
\(-2x^2+5x-16=0\)
\(-\left(2x^2-5x+16\right)=0\)
\(2x^2-5x+16=0\)
\(2\left(x^2-\frac{5}{2}x+8\right)=0\)
\(x^2-\frac{5}{2}x+8=0\)
\(x^2-\frac{5}{2}x+\frac{25}{16}+\frac{103}{16}=0\)
\(\left(x-\frac{5}{4}\right)^2+\frac{103}{16}=0\)
Ta có: \(\left(x-\frac{5}{4}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-\frac{5}{4}\right)^2+\frac{103}{16}\ge\frac{103}{16}>0\)
Mà: \(\left(x-\frac{5}{4}\right)^2+\frac{103}{16}=0\)
=> Vô lí
Vậy : ko có giá trị thỏa mãn của x
=.= hok tốt!!
\(x^2+2x-99=0\)
\(x^2+2x+1-100=0\)
\(\left(x+1\right)^2-10^2=0\)
\(\left(x+1+10\right)\left(x+1-10\right)=0\)
\(\left(x+11\right)\left(x-9\right)=0\)
\(\left[\begin{array}{l}x+11=0\Rightarrow x=-11\\ x-9=0\Rightarrow x=9\end{array}\right.\)
KL: x = -11; x = 9
x^2+2x-99=0
x^2+2x=99
x(x+2)=99
x(x+2)=9x11
x=9
Vậy x=9