Sửa đề: \(\dfrac{x+1}{x-2}+\dfrac{1}{x+2}=\dfrac{5x-4}{x^2-4}\)
=>x^2+3x+2+x-2=5x-4
=>x^2+4x=5x-4
=>x^2-x+4=0
Δ=(-1)^2-4*1*4=1-16=-15<0
=>PTVN
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1. Rút Gọn
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Thanks mn nhiều ạ
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14 tháng 3 2022 lúc 10:06
1) (3x-2)/3-2=(4x+1)/42) (x-3)/4+(2x-1)/3=(2-x)/63) 1/2 (x+1)+1/4 (x+3)=3-1/3 (x+2)4) (x+4)/5-x+4=x/3-(x-2)/25) (4-5x)/6=2(-x+1)/2 6) (-(x-3))/2-2=5(x+2)/4 7)2(2x+1)/5-(6+x)/3=(5-4x)/158) (7-3x)/2-(5+x)/5=1 9)(x-1)/2+3(x+1)/8=(11-5x)/310)(3+5x)/5-3=(9x-3)/4
a. P=(5x−1)2+2(1−5x)(4+5x)+(5x+4)2P=(5x−1)2+2(1−5x)(4+5x)+(5x+4)2
b. Q=(x−y)3+(y+x)3+(y−x)3−3xy(x+y)
1) 2x – (3 – 5x) = 4( x +3)
2) 5(2x-3) - 4(5x-7) =19 - 2(x+11)
3) 5x - 4(6-x)(x + 3) = (4-2x)(3-2x) + 2
4) (x - 1)3 - (3x + 2)(-12) = (x2 + 1)(x - 2) - x2
5) (3x -1)2 - (x +3)(2x-1) = 7(x + 1)(x -2) -3x
mn giúp mình vs
(x-1)(x-2)(x-3)(x-4)-80chia hết cho x^2-5x
và x^8-2x^5-2x^4+x^2-2x-100+10x(x^4+x)+(5x-1)^2chia hết cho x^2-5x-4
tìm x biếta,2x22+3(x-1)(x+1)5x(x+1)b,,(8-5x)(x+2)+4(x-2)(x+1)(2+x)(2-x)c,, 4(x-1)(x+5)-(x+2)(x+5)3(x-1)(x+2)
Đọc tiếp
tìm x biết
a,2x+3(x-1)(x+1)=5x(x+1)
b,,(8-5x)(x+2)+4(x-2)(x+1)=(2+x)(2-x)
c,, 4(x-1)(x+5)-(x+2)(x+5)=3(x-1)(x+2)
phân tích đa thức
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