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4 tháng 7 2017
a) ( 3x - 1 ) ( 2x + 7 ) - ( x + 1 ) ( 6x + 5 ) = 16
<=> 6x2 + 21x - 2x - 7 - ( 6x2 - 5x + 6x - 5) = 16
<=> 6x2 + 21x - 2x - 7 - ( 6x2 + x - 5 ) = 16
<=> 6x2+ 21x - 2x - 7 - 6x2 -x + 5 = 16
<=> 18x - 2 = 16
<=> 18x = 18
=> x = 1
Vậy....
PN
1
TM
22 tháng 7 2016
a)\(x\left(x+2\right)-3x-6=0\)
=>\(x\left(x+2\right)-3\left(x+2\right)=0\)
=>\(\left(x-3\right)\left(x+2\right)=0\)
=>\(\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
b)\(x^3+3x^2+3x-1-3x^2-3x=0\)
=>\(x^3-1=0\)
=>x3=1
=>x=1
18 tháng 12 2016
a) \(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(\Leftrightarrow2x^2-10x-3x-2x^2=26\)
\(\Leftrightarrow-13x=26\Leftrightarrow x=-2\)
b) \(5x\left(x-1\right)=x-1\)
\(\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=\frac{1}{5}\end{array}\right.\)
c) \(2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-5\\x=2\end{array}\right.\)
d) \(\left(2x-3\right)^2-\left(x+5\right)^2=0\)
\(\Leftrightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\)
\(\Leftrightarrow\left(x-8\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=8\\x=-\frac{2}{3}\end{array}\right.\)
e) \(3x^3-48x=0\)
\(\Leftrightarrow3x\left(x^2-16\right)=0\)
\(\Leftrightarrow3x\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=4\\x=-4\end{array}\right.\)
f) \(x^3+x^2-4x=4\)
\(\Leftrightarrow x^2\left(x+1\right)-4\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=2\\x=-2\end{array}\right.\)
24 tháng 3 2020
a) 2(x + 5) - x^2 - 5x = 0
<=> 2x + 10 - x^2 - 5x = 0
<=> -3x + 10 - x^2 = 0
<=> x^2 + 3x - 10 = 0
<=> (x - 2)(x + 5) = 0
<=> x - 2 = 0 hoặc x + 5 = 0
<=> x = 2 hoặc x = -5
b) 2(x - 3)(x^2 + 1) + 15x - 5x^2 = 0
<=> 2x^3 + 2x - 6x^2 - 6 + 15x - 5x^2 = 0
<=> 2x^3 + 17x - 11x^2 - 6 = 0
<=> (2x^2 - 7x + 3)(x - 2) = 0
<=> (2x^2 - x - 6x + 3)(x - 2) = 0
<=> [x(2x - 1) - 3(2x - 1)](x - 2) = 0
<=> (x - 3)(2x - 1)(x - 2) = 0
<=> x - 3 = 0 hoặc 2x - 1 = 0 hoặc x - 2 = 0
<=> x = 3 hoặc x = 1/2 hoặc x = 2
c) (x + 2)(3 - 4x) = x^2 + 4x + 2
<=> 3x - 4x^2 + 6 - 8x = x^2 + 4x + 2
<=> -5x - 4x^2 + 6 = x^2 + 4x + 2
<=> 5x + 4x^2 - 6 + x^2 + 4x + 2 = 0
<=> 9x + 5x^2 - 4 = 0
<=> 5x^2 + 10x - x - 4 = 0
<=> 5x(x + 2) - (x + 2) = 0
<=> (5x - 1)(x + 2) = 0
<=> 5x - 1 = 0 hoặc x + 2 = 0
<=> x = 1/5 hoặc x = -2
5 tháng 8 2016
3) \(\left(x-1\right)\left(x+1\right)^2-\left(2x-1\right)\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)^2-\left(2x-1\right)\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)^2\left(x-1-2x+1\right)=0\)
\(\Leftrightarrow-x\left(x+1\right)^2=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}-x=0\\\left(x+1\right)^2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x+1=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=-1\end{array}\right.\)
4 tháng 7 2017
Bài 1
Em xem lại đề nhé
a. Ta có VP=\(x^4-y^4=\left(x^2\right)^2-\left(y^2\right)^2\)
\(=\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x+y\right)\left(x-y\right)\left(x^2+y^2\right)\)
\(=\left(x+y\right)\left(x^3+xy^2-x^2y-y^3\right)\)
\(=VT\)
b.
1.\(\left(x-3\right)\left(x-2\right)-\left(x+10\right)\left(x-5\right)=0\)
\(\Leftrightarrow x^2-5x+6-\left(x^2+5x-50\right)=0\)
\(\Leftrightarrow-10x=-56\Rightarrow x=\frac{56}{10}\)
2.\(\left(2x-1\right)\left(3-x\right)+\left(x-2\right)\left(x+3\right)=\left(1-x\right)\left(x-2\right)\)
\(=-2x^2+7x-3+x^2+x-6=-x^2+3x-2\)
\(\Leftrightarrow5x=7\Leftrightarrow x=\frac{7}{5}\)
AH
Akai Haruma
Giáo viên
28 tháng 7 2018
Lời giải:
\(P(x)=x(x+2)(x+3)(x+5)-7\)
\(=[x(x+5)][(x+2)(x+3)]-7\)
\(=(x^2+5x)(x^2+5x+6)-7\)
\(=a(a+6)-7\) (đặt \(x^2+5x=a\) )
\(=a^2+6a-7=a^2-a+7a-7\)
\(=a(a-1)+7(a-1)=(a-1)(a+7)\)
\(=(x^2+5x-1)(x^2+5x+7)\)
-----------------
\(Q(x)=(4x-2)(10x+4)(5x+7)(2x+1)+17\)
\(=4(2x-1)(5x+2)(5x+7)(2x+1)+17\)
\(=4[(2x-1)(5x+7)][(5x+2)(2x+1)]+17\)
\(=4(10x^2+9x-7)(10x^2+9x+2)+17\)
\(=4a(a+9)+17\) (đặt \(10x^2+9x-7=a\)
\(=4a^2+36a+17=(2a+9)^2-8^2\)
\(=(2a+9-8)(2a+9+8)=(2a+1)(2a+17)\)
\(=(20x^2+18x-13)(20x^2+18x+3)\)
AH
Akai Haruma
Giáo viên
28 tháng 7 2018
\(R(x)=(3x+2)(3x-5)(x-1)(9x+10)+24x^2\)
\(=[(3x+2)(3x-5)][(x-1)(9x+10)]+24x^2\)
\(=(9x^2-9x-10)(9x^2+x-10)+24x^2\)
\(=(a-9x)(a+x)+24x^2\) (đặt \(9x^2-10=a\) )
\(=a^2-8ax+15x^2=(a^2-5ax)-(3ax-15x^2)\)
\(=a(a-5x)-3x(a-5x)=(a-3x)(a-5x)\)
\(=(9x^2-3x-10)(9x^2-5x-10)\)
--------------------------
\(H(x)=(x-18)(x-7)(x+35)(x+90)-67x^2\)
\(=[(x-18)(x+35)][(x-7)(x+90)]-67x^2\)
\(=(x^2+17x-630)(x^2+83x-630)-67x^2\)
\(=a(a+66x)-67x^2\) (đặt \(x^2+17x-630=a\) )
\(=a^2-ax+67ax-67x^2\)
\(=a(a-x)+67x(a-x)=(a-x)(a+67x)\)
\(=(x^2+16x-630)(x^2+84x-630)\)
a) TH1 : \(x-1=0\)
\(\Rightarrow x=1\)
TH2 : \(x-1\ne0\)
\(\Rightarrow5x\left(x-1\right)=1.\left(x-1\right)\)
\(5x=1\)
\(x=\frac{1}{5}\)
Vậy ...
b) \(2\left(x+5\right)-x^2-5x=0\)
\(2\left(x+5\right)-\left(x^2+5x\right)=0\)
\(2\left(x+5\right)-x\left(x+5\right)=0\)
\(\left(2-x\right)\left(x+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2-x=0\\x+5=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)
a) 5x(x - 1) = x - 1
=> 5x(x - 1)
b) 2(x + 5) - x2 - 5x = 0
2(x + 5) + (-x2 - 5x) = 0
=> 2(x + 5) - x(x + 5) = 0
=> (x + 5) (2 - x) = 0
=> x + 5 = 0 => x = -5
=> 2 - x = 0 => x = 2
t i c k nhé!! 45345345366454676576878708673454255135454365464564756