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a: \(\dfrac{1}{x-2}+3=\dfrac{3-x}{x-2}\)
=>1+3x-6=3-x
=>3x-5=3-x
=>4x=8
hay x=2(loại)
b: \(\Leftrightarrow8-x-8\left(x-7\right)=-26\)
=>8-x-8x+56=-26
=>-9x+64=-26
=>-9x=-90
hay x=10(nhận)
c: \(\dfrac{1}{x-2}+\dfrac{1}{x-3}=\dfrac{2}{x-1}\)
\(\Leftrightarrow\dfrac{x-3+x-2}{\left(x-2\right)\left(x-3\right)}=\dfrac{2}{x-1}\)
\(\Leftrightarrow\left(x-1\right)\left(2x-5\right)=2\left(x^2-5x+6\right)\)
\(\Leftrightarrow2x^2-5x-2x+5=2x^2-10x+12\)
=>-7x+10x=12-5
=>3x=7
hay x=7/3(nhận)
bÀI LÀM
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
a) 4x2 - 2x + 3 - 4x.(x - 5) = 7x - 3
--> 4x2 - 2x + 3 - 4x2 + 20x = 7x - 3
--> 4x2 - 2x - 4x2 + 20x - 7x = -3 - 3
--> 11x = -6
--> x = \(\frac{-6}{11}\)
b) -3x.(x - 5) + 5.(x - 1) + 3x2 = 4x
--> -3x2 + 15x + 5x - 5 + 3x2 = 4x
--> -3x2 + 15x + 5x + 3x2 - 4x = 5
--> 16x = 5
--> x = \(\frac{5}{16}\)
c) 7x.(x - 2) - 5.(x - 1) = 21x2 - 14x2 + 3
--> 7x2 - 14x - 5x + 5 = 7x2 + 3
--> 7x2 - 14x - 5x - 7x2 = -5 + 3
--> -19x = -2
--> x = \(\frac{2}{19}\)
d) 3.(5x - 1) - x.(x - 2) + x2 - 13x = 7
--> 15x - 3 - x2 + 2x + x2 - 13x = 7
--> 15x - x2 + 2x + x2 - 13x = 3 + 7
--> 4x = 10
--> x = \(\frac{5}{2}\)
e) \(\frac{1}{5}\)x.(10x - 15) - 2x.(x - 5) = 12
--> 2x2 - 3x - 2x2 + 10x = 12
--> 7x = 12
--> x = \(\frac{12}{7}\)
~ Học tốt ~
a) 4x2 - 2x + 3 - 4x(x - 5) = 7x - 3
=> 4x2 - 2x + 3 - 4x2 + 20x = 7x - 3
=> 18x + 3 = 7x - 3
=> 18x - 7x = -3 - 3
=> 11x = -6
=> x = -6/11
b) -3x(x - 5) + 5(x - 1) + 3x2 = 4x
=> -3x2 + 15x + 5x - 5 + 3x2 = 4x
=> 20x - 5 = 4x
=> 20x - 4x = 5
=> 16x = 5
=> x = 5/16
\(c,7x\left(x-2\right)-5\left(x-1\right)=21x^2-14x^2+3\)
\(\Leftrightarrow7x^2-14x-5x+5=7x^2+3\)
\(\Leftrightarrow7x^2-7x^2-19x=3-5\)
\(\Leftrightarrow-19x=-2\)
\(\Leftrightarrow x=\frac{2}{19}\)
a, ( x2 + x )2 - 14 ( x2 + x ) + 24
= (x2 + x)2 - 2(x2 + x) -12(x2 + x) + 24
= (x2 + x).(x2 + x -2) - 12(x2 + x -2)
= (x2 + x -2).(x2 + x -12)
= (x2 + 2x - x - 2).(x2 + 4x - 3x - 12)
=[x.(x+2)-(x+2)].[x.(x+4)-3(x+4)]
= (x+2).(x-1).(x+4).(x-3)
= x4 + 2x3 - 13x2 - 14x + 24
b, ( x2 + x )2 + 4x2 + 4x - 12
= x4 + 2x3 + x2 + 4x2 + 4x -12
= x4 + 2x3 + 5x2 + 4x -12
c, x4 + 2x3 + 5x2 + 4x - 12
= x4 - x3 + 3x3 - 3x2 + 8x2 - 8x +12x -12
= x3(x-1) + 3x2(x-1) + 8x(x-1) + 12(x-1)
= (x-1) . (x3 + 3x2 + 8x +12)
= (x-1) . ( x3 +2x2 + x2 + 2x + 6x +12)
= (x-1). [x2(x+2) + x(x+2) + 6(x+2)]
= (x-1).(x+2).(x2 + x+ 6)
b) \(x^5+x+1=x^5-x^2+x^2+x+1=x^2\left(x^3-1\right)+x^2+x+1=x^2\left(x-1\right)\left(x^2+x+1\right)+x^2+x+1=\left(x^2+x+1\right)\left[x^2\left(x-1\right)+1\right]\)