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Bài 1 :
a, \(\left(a-2\right)^2-b^2=\left(a-2-b\right)\left(a-2+b\right)\)
b, \(2a^3-54b^3=2\left(a^3-27b^3\right)=2\left(a-3b\right)\left(a^2+3ab+9b\right)\)
Bài 2 : tự kết luận nhé, ngại mà lười :(
a, \(\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\)
\(\Leftrightarrow\frac{4x-3}{5}-\frac{5x-4}{3}=\frac{6x-2}{7}+3\)
\(\Leftrightarrow\frac{12x-9-25x+20}{15}=\frac{6x-2+21}{7}\)
\(\Leftrightarrow\frac{-13x-29}{15}=\frac{6x+19}{7}\Rightarrow-91x-203=90x+285\)
\(\Leftrightarrow181x=-488\Leftrightarrow x=-\frac{488}{181}\)
b, \(\frac{x+2}{3}+\frac{3\left(2x-1\right)}{4}-\frac{5x-3}{6}=x+\frac{5}{12}\)
\(\Leftrightarrow\frac{4x+8+9\left(2x-1\right)}{12}-\frac{10x-6}{12}=\frac{12x+5}{12}\)
\(\Rightarrow4x+8+18x-9-10x+6=12x+5\)
\(\Leftrightarrow12x+5=12x+5\Leftrightarrow0x=0\)
Vậy phương trình có vô số nghiệm
c, \(\left|2x-3\right|=4\)
Với \(x\ge\frac{3}{2}\)pt có dạng : \(2x-3=4\Leftrightarrow x=\frac{7}{2}\)
Với \(x< \frac{3}{2}\)pt có dạng : \(2x-3=-4\Leftrightarrow x=-\frac{1}{2}\)
d, \(\left|3x-1\right|-x=2\Leftrightarrow\left|3x-1\right|=x+2\)
Với \(x\ge\frac{1}{3}\)pt có dạng : \(3x-1=x+2\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)
Với \(x< \frac{1}{3}\)pt có dạng : \(3x-1=-x-2\Leftrightarrow4x=-1\Leftrightarrow x=-\frac{1}{4}\)
Vì dài quá nên mình chỉ có thể trả lời được mấy câu thôi
Bài 1:
27x3 - 8 : (6x + 9x2 +4)
= (3x - 2) (9x2 + 6x + 4) : (9x2 + 6x + 4)
= 3x - 2
Bài 3:
a, 81x4 + 4 = (9x2)2 + 36x2 + 4 - 36x2
= (9x2 + 2)2 - (6x)2
= (9x2 + 6x + 2)(9x2 - 6x + 2)
b, x2 + 8x + 15 = x2 + 3x + 5x + 15
= x(x + 3) + 5(x + 3)
= (x + 3)(x + 5)
c, x2 - x - 12 = x2 + 3x - 4x - 12
= x(x + 3) - 4(x + 3)
= (x + 3) (x - 4)
Câu 1:
(27x3 - 8) : (6x + 9x2 + 4)
= (3x - 2)(9x2 + 6x + 4) : (6x + 9x2 + 4)
= 3x - 2
Câu 2:
a) (3x - 5)(2x+ 11) - (2x + 3)(3x + 7)
= 6x2 + 33x - 10x - 55 - 6x2 - 14x - 9x - 21
= -76
⇒ đccm
b) (2x + 3)(4x2 - 6x + 9) - 2(4x3 - 1)
= 8x3 + 27 - 8x3 + 2
= 29
⇒ đccm
Câu 3:
a) 81x4 + 4
= (9x2)2 + 22
= (9x2 + 2)2 - (6x)2
= (9x2 - 6x + 2)(9x2 + 6x + 2)
b) x2 + 8x + 15
= x2 + 3x + 5x + 15
= x(x + 3) + 5(x + 3)
= (x + 3)(x + 5)
c) x2 - x - 12
= x2 - 4x + 3x - 12
= x(x - 4) + 3(x - 4)
= (x - 4)(x + 3)
a) ĐKXĐ: x khác 0
\(x+\dfrac{5}{x}>0\)
\(\Leftrightarrow x^2+5>0\) ( luôn đúng)
Vậy bất pt vô số nghiệm ( loại x = 0)
d)
\(\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2}{8}-\dfrac{x+3}{8}\)
\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2-x-3}{8}\)
\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{-5}{8}\)
\(\Leftrightarrow2x+2-4x+4>-15\)
\(\Leftrightarrow-2x>-21\)
\(\Leftrightarrow x< \dfrac{21}{2}\)
Vậy....................
a)\(x+\dfrac{5}{x}>0\left(ĐKXĐ:x\ne0\right)\)
\(\Leftrightarrow\dfrac{x^2+5}{x}>0\)
Mà \(x^2+5>0\)
\(\Rightarrow x>0\)
d)\(\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2}{8}-\dfrac{x+3}{8}\)
\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{2x-2}{12}>\dfrac{-5}{8}\)
\(\Leftrightarrow\dfrac{-x+3}{12}>\dfrac{-5}{8}\)
\(\Leftrightarrow-x+3>-\dfrac{15}{2}\)
\(\Leftrightarrow-x>-\dfrac{21}{2}\)
\(\Leftrightarrow x< \dfrac{21}{2}\)
Bài 2:
\(=\dfrac{x^2\left(x^2+4\right)-2x\left(x^2+4\right)}{x^2+4}=x^2-2x\)
Bài 1:
a: \(=\left(\dfrac{2}{3}:\dfrac{-1}{9}\right)\cdot x^4y^2z^6=-6x^4y^2z^6\)
b: \(=-12x^8-21x^5\)
c: =x^3+8
d: \(=125x^3-75x^2+15x-1\)
a) \(\left(6x^3y^2-4x^2y^3-10x^2y^2\right):2xy\)
=\(\left(6x^3y^2:2xy\right)-\left(4x^2y^3:2xy\right)-\left(10x^2y^2:2xy\right)\)
\(=3x^2y-2xy^2-5xy\)
b) \(\dfrac{2y}{x-2}+\dfrac{5y}{x-2}\)
=\(\dfrac{2y+5y}{x-2}\)
=\(\dfrac{7y}{x-2}\)
c)\(\dfrac{xy}{3x-y}+\dfrac{3x^2}{y-3x}\)
\(=\dfrac{xy}{3x-y}-\dfrac{3x^2}{3x-y}\)
=\(\dfrac{x\left(y-3x\right)}{3x-y}\)
=\(\dfrac{-x\left(3x-y\right)}{3x-y}\)
=-x
d)\(\dfrac{x-1}{6x+12}.\dfrac{x+2}{x-1}\)
=\(\dfrac{\left(x-1\right)\left(x+2\right)}{6\left(x+2\right)\left(x-1\right)}\)
=\(\dfrac{1}{6}\)
bài 1:
b,\(\dfrac{x+2}{x}=\dfrac{x^2+5x+4}{x^2+2x}+\dfrac{x}{x+2}\)(ĐKXĐ:x ≠0,x≠-2)
<=>\(\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}=\dfrac{x^2+5x+4}{x\left(x+2\right)}+\dfrac{x^2}{x\left(x+2\right)}\)
=>\(x^2+4x+4=x^2+5x+4+x^2\)
<=>\(x^2-x^2-x^2+4x-5x+4-4=0\)
<=>\(-x^2-x=0< =>-x\left(x+1\right)=0< =>\left[{}\begin{matrix}x=0\left(loại\right)\\x+1=0< =>x=-1\left(nhận\right)\end{matrix}\right.\)
vậy...............
d,\(\left(x+3\right)^2-25=0< =>\left(x+3-5\right)\left(x+3+5\right)=0< =>\left(x-2\right)\left(x+8\right)=0< =>\left[{}\begin{matrix}x-2=0\\x+8=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=2\\x=-8\end{matrix}\right.\)
vậy............
bài 3:
g,\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x^2-x-2}\)(ĐKXĐ:x khác -1,x khác 2)
<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x^2-2x+x-2}\)
<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x\left(x-2\right)+\left(x-2\right)}\)
<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{\left(x+1\right)\left(x-2\right)}\)
<=>\(\dfrac{4\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}-\dfrac{2\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{x+3}{\left(x+1\right)\left(x-2\right)}\)
=>\(4x-8-2x-2=x+3\)
<=>\(x=13\)
vậy..............
mấy ý khác bạn làm tương tụ nhé
chúc bạn học tốt ^ ^
a) 5x - 15y = 5(x - 3y)
b) \(\dfrac{3}{5}\)x2 + 5x4 - x2 - y
= \(\dfrac{3}{5}\)x2 + 5x2.x2 - x2 - y
= x2(\(\dfrac{3}{5}\) + 5x2 -1) - y
c) 14x2y2 - 21xy2 + 28x2y
= 7xy.xy - 7xy.3y + 7xy.4x
= 7xy(xy - 3y + 4x)
= 7xy[(xy - 3y) + 4x]
= 7xy[y(x - 3) +4x]
d) \(\dfrac{2}{7}x\)(3y - 1) - \(\dfrac{2}{7}y\)(3y - 1)
= (3y - 1).(\(\dfrac{2}{7}x\) - \(\dfrac{2}{7}y\) )
= (3y - 1).[\(\dfrac{2}{7}\)(x - y)]
e) x3 - 3x2 + 3x - 1
= x2.x - 3x.x + 3.x - 1
= x(x2-3x+3) - 1
g) 27x3 + \(\dfrac{1}{8}\)
= (3x)3 + \(\left(\dfrac{1}{2}\right)^3\)
= (3x + \(\dfrac{1}{2}\)).(9x2 - \(\dfrac{3}{2}\)x + \(\dfrac{1}{4}\))
h) (x+y)3 - (x-y)3
= 2(3x2y) + 2y3
f) (x+y)2 - 4x2
= -3x2 + y(2x + y)
Câu 1:
a: =(y-3)(x^2-16)
=(x-4)(x+4)(y-3)
b: \(=\left(2x+1\right)^2-y^2\)
\(=\left(2x+1+y\right)\left(2x+1-y\right)\)
Câu 1:
\(\left(x^2+2x+4\right)\left(x-2\right)=x^3-8\)
Câu 3:
a) \(\dfrac{7x-4}{x-1}-\dfrac{5x-2}{x-1}=\dfrac{7x-4-5x+2}{x-1}=\dfrac{2\left(x-1\right)}{x-1}=2\)
b) \(\dfrac{10x}{27y^2}:\dfrac{5x^3}{9xy}=\dfrac{10x}{27y^2}\cdot\dfrac{9xy}{5x^3}=\dfrac{90x^2y}{135x^3y^2}=\dfrac{2}{3xy}\)
Câu 4:
a) \(x^2\left(x-3\right)-4x+12\)
\(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-4\right)\)
\(=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
b) \(a^2-2ab+b^2-9\)
\(=\left(a-b\right)^2-3^2\)
\(=\left(a-b-3\right)\left(a-b+3\right)\)
c) \(6x^3+18x^2y+18xy^2+6y^3\)
\(=6\left(x^3+3x^2y+3xy^2+y^3\right)\)
\(=6\left(x+y\right)^3\)
Bài 5:
a) Phân thức P có giá trị xác định khi
\(x^2-3x\ne0\Rightarrow x\left(x-3\right)\ne0\)
\(\Rightarrow\left\{{}\begin{matrix}x\ne0\\x-3\ne0\Rightarrow x\ne3\end{matrix}\right.\)
b) \(P=\dfrac{4x-12}{x^2-3x}=\dfrac{4\left(x-3\right)}{x\left(x-3\right)}=\dfrac{4}{x}\)
Với \(x\ne0,x\ne3\) thì P = 4
\(\Rightarrow\dfrac{4}{x}=4\Rightarrow x=16\)
Câu 1:Tính
\(\left(x^2+2x+4\right).\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+2x+4\right)\)
\(=x^3-8\)
Câu 2:tính
a,\(\dfrac{7x-4}{x-1}-\dfrac{5x-2}{x-1}=\dfrac{7x-4-5x+2}{x-1}=\dfrac{2x-2}{x-1}=\dfrac{2\left(x-1\right)}{x-1}=2\)
\(b,\dfrac{10x}{27y^2}:\dfrac{5x^3}{9xy}=\dfrac{10x}{27y^2}.\dfrac{9xy}{5x^3}=\dfrac{2}{3xy}\)
Câu 3:Phân tích các đa thức sau thành nhân tử
\(a,x^2\left(x-3\right)-4x+12\)
\(=x^2\left(x-3\right)-\left(4x-12\right)\)
\(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-4\right)\)
\(=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
\(b,a^2-2ab+b^2-9\)
\(=\left(a^2-2ab+b^2\right)-3^2\)
\(=\left(a-b\right)^2-3^2\)
\(=\left(a-b-3\right)\left(a-b+3\right)\)
\(c,6x^3+18x^2y+18xy^2+6y^3\)
\(=6\left(x^3+3x^2y+3xy^2+y^3\right)\)
\(=6\left(x+y\right)^3\)
Còn câu cuối lát mk lm nốt , bh mk bận nha bn .
Câu 5 :
a, ĐKXĐ của P là :\(x^2-3x\ne0\Leftrightarrow x\left(x-3\right)\ne0\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne3\end{matrix}\right.\)
b, \(P=\dfrac{4x-12}{x^2-3x}=\dfrac{4\left(x-3\right)}{x\left(x-3\right)}=\dfrac{4}{x}\)
Thay x = 4 vào biểu thức P ,có :
\(\dfrac{4}{4}=1\)
Vậy tại x=4 giá trị của biểu thức P là 1
c, Để P nguyên
⇔ \(\dfrac{4}{x}\) nguyên
\(\Leftrightarrow4⋮x\)
\(\Leftrightarrow x\inƯ\left(4\right)=\left\{1;-1;4;-4;2;-2\right\}\)( t/m ĐKXĐ )
Vậy \(x\in\left\{1;-1;4;-4;2;-2\right\}\)
thì P nguyên
1.
\(\left(x^2+2x+4\right)\left(x-2\right)\\ =x^3-8\)
2.
a.
\(\dfrac{7x-4}{x-1}-\dfrac{5x-2}{x-1}=\dfrac{2x-2}{x-1}=2\)
b. Ko khó
3.
\(x^2\left(x-3\right)-4x+12\\ =x^2\left(x-3\right)-4\left(x-3\right)\\ =\left(x-3\right)\left(x^2-4\right)\\ =\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
b.
\(a^2-2ab+b^2-9\\ =\left(a-b\right)^2-3^2\\ =\left(a-b-3\right)\left(a-b+3\right)\)
5.
P xác định <=> x^2-3x khác 0 <=> x khác 0 ; x khác 3
b.
\(P=\dfrac{4x-12}{x^2-3x}=\dfrac{4\left(x-3\right)}{x\left(x-3\right)}=\dfrac{4}{x}\)
P=4 <=>x=1
c.
4 chia hết x
Câu b mk nhầm x = 4 nha bn
b, Để P = 4
⇔ \(\dfrac{4}{x}=4\Leftrightarrow x=1\)
C5 ; x= 1
Ngô Tấn Đạt
uk
nhầm
lại đi lấy 4*4
hihi
thank you so much
thank you so so much
cảm ơn bạn nhiều nha !!!!!!!!bạn có thể giúp mik bài hình được không
bài hình nào bn