Bài 4: Chứng minh rằng các đẳng thức sau bằng nhau
a)\(\dfrac{3x\left(x+5\right)}{2\left(x+5\right)}\)=\(\dfrac{6x^2+30x}{4}\)
b)\(\dfrac{x+2}{x-1}\)=\(\dfrac{\left(x+2\right)\left(x+1\right)}{x^2-1}\)
a/ ĐK: $x\ne -5$
$\dfrac{6x^2+30x}{4}=\dfrac{6x(x+5)}{4}=\dfrac{3x(x+5)}{2}$
Đề này sai
b/ ĐK: $x\ne \pm 1$
$\dfrac{(x+2)(x+1)}{x^2-1}\\=\dfrac{(x+2)(x+1)}{(x-1)(x+1)}\\=\dfrac{x+2}{x-1}$
$\to$ ĐPCM
a, Xét \(VT=\dfrac{3x\left(x+5\right)}{2\left(x+5\right)}=\dfrac{3x}{2}\)
\(VP=\dfrac{6x^2+30x}{4}=\dfrac{6x\left(x+5\right)}{4}=\dfrac{3x\left(x+5\right)}{2}\)
Vậy \(VT\ne VP\)hay đpcm ko xảy ra
b, \(VP=\dfrac{\left(x+2\right)\left(x+1\right)}{x^2-1}=\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+2}{x-1}=VT\)
Vậy ta có đpcm
Cho A = \(\dfrac{\left(x-y\right)^2+xy}{\left(x+y\right)^2-xy}.\left[1:\dfrac{x^5+y^5+x^3y^2+x^2y^3}{\left(x^3-y^3\right)\left(x^3+y^3+x^2y+xy^2\right)}\right]\)
B = x - y
Chứng minh đẳng thức A = B
Tính giá trị của A, B tại x = 0; y = 0 và giải thích vì sao A ≠ B
\(ĐK:x\ne y;x\ne-y;x^2+xy+y^2\ne0;x^2-xy+y^2\ne0\)
\(A=\dfrac{x^2-xy+y^2}{x^2+xy+y^2}\cdot\left[1:\dfrac{\left(x^3+y^3\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2+y^2\right)}\right]\\ A=\dfrac{x^2-xy+y^2}{x^2+xy+y^2}\cdot\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)\left(x^2+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)\left(x^2+y^2\right)}\\ A=x-y=B\)
\(x=0;y=0\Leftrightarrow B=0\)
Giá trị của A không xác định vì \(x=y\) trái với ĐK:\(x\ne y\)
Vậy \(A\ne B\)
Dùng định nghĩa hai phân thức bằng nhau chứng tỏ :
a) \(\dfrac{5y}{7}=\dfrac{20xy}{28x}\)
b) \(\dfrac{3x\left(x+5\right)}{2\left(x+5\right)}=\dfrac{3x}{2}\)
c) \(\dfrac{x+2}{x-1}=\dfrac{\left(x+2\right)\left(x+1\right)}{x^2-1}\)
d) \(\dfrac{x^2-x-2}{x+1}=\dfrac{x^2-3x+2}{x-1}\)
Bài 1: (Sgk/36):
a. \(\dfrac{5y}{7}\)=\(\dfrac{20xy}{28x}\) vì
5y . 28x = 140xy
7 . 20xy = 140xy
=> 5y . 28x = 7 . 20xy
Vậy \(\dfrac{5y}{7}\)=\(\dfrac{20xy}{28x}\)
b. \(\dfrac{3x\left(x+5\right)}{2\left(x+5\right)}\)=\(\dfrac{3x}{2}\) vì
3x . 2(x+5) = 6x2+30x
2 . 3x(x+5) = 6x2+30x
=> 3x . 2(x+5) = 2 . 3x(x+5)
Vậy \(\dfrac{3x\left(x+5\right)}{2\left(x+5\right)}\)=\(\dfrac{3x}{2}\)
c. \(\dfrac{x+2}{x-1}\)=\(\dfrac{\left(x+2\right)\left(x+1\right)}{x^2-1}\) vì
(x+2) (x2-1) = (x+2) (x-1) (x-1)
=> (x+2) (x2-1) = (x-1) (x+2) (x+1)
Vậy \(\dfrac{x+2}{x-1}\)=\(\dfrac{\left(x+2\right)\left(x+1\right)}{x^2-1}\)
d. \(\dfrac{x^2-x-2}{x+1}\)=\(\dfrac{x^2-3x+2}{x-1}\)
(x-1) (x2-x-2) = x3-2x2-x+2
(x+1) (x2-3x+2) = x3-2x2-x+2
=> (x-1) (x2-x-2) = (x2-3x+2) (x+1)
Vậy \(\dfrac{x^2-x-2}{x+1}\)=\(\dfrac{x^2-3x+2}{x-1}\)
chứng minh đẳng thức sau
a. \(\dfrac{3y}{4}=\dfrac{6xy}{8x}\left(x\ne0\right)\)
b. \(\dfrac{x+y}{3a}=\dfrac{3a\left(x+y\right)^2}{9a^2\left(x+y\right)}\)
\(a,VT=\dfrac{3y\cdot2x}{4\cdot2x}=\dfrac{6xy}{8x}=VP\\ b,VT=\dfrac{\left(x+y\right)\cdot3a\left(x+y\right)}{3a\cdot3a\left(x+y\right)}=\dfrac{3a\left(x+y\right)^2}{9a^2\left(x+y\right)}=VP\)
C/minh các đẳng thức sau:
\(a,\dfrac{2\left(x-y\right)}{3y-3x}=\dfrac{-2}{3}\)
\(b,\dfrac{x-2}{-x}=\dfrac{8-x^3}{x\left(x^2+2x+4\right)}\)
\(c,\dfrac{3x}{x+y}=\dfrac{-3x\left(x-y\right)}{y^2-x^2}\)
a)2(x-y)/(-3)(x-y)=-2/3
b)8-x^3=(2-x)(x^2+2x+4) => Vế phải =(2-x)/x=(x-2)/-x
c)y^2-x^2=(y+x)(y-x) bạn đổi dấu rồi rút gọn là được,cũng tương tự như trên ý
bài 1. rút gọn phân thức sau :
a, \(\dfrac{15xy}{10x^2y}\)
b, \(\dfrac{4\left(2-x\right)}{6x\left(x-2\right)}\)
c, \(\dfrac{12x^3\left(x-4\right)^2}{8x^2\left(4-x\right)}\)
d, \(\dfrac{6x\left(x+5\right)^3}{2x^2\left(x+5\right)}\)
bài 2. rút gọn phân thức sau :
a, \(\dfrac{15\left(x-3\right)^3}{9-3x}\)
b, \(\dfrac{4\left(5x-1\right)^3}{2-10x}\)
c,\(\dfrac{9x^2y}{-15x^3y}\)
d, \(\dfrac{-15x\left(x-y\right)^2}{6x^2\left(x-y\right)^3}\)
bài 3. rút gọn phân thức sau :
a, \(\dfrac{5x^2+10x+5}{11x+11}\)
b, \(\dfrac{18x+6x^2+2x^3}{x^3-27}\)
c, \(\dfrac{12x^2+12x+3}{4x^2-1}\)
bài 4 : rút gọ phân thức đại số
A= \(\dfrac{12xy^2}{-9x^3y}\)
B=\(\dfrac{35xy\left(x-4\right)^2}{28x^2\left(4-x\right)}\)
C= \(\dfrac{6x+18y}{x^2-9y^2}\)
Bài 1:
a) \(\dfrac{15xy}{10x^2y}\)
= \(\dfrac{3.5xy}{2.5xyx}\)
= \(\dfrac{3}{2x}\)
d) \(\dfrac{6x\left(x+5\right)^3}{2x^2\left(x+5\right)}\)
= \(\dfrac{3.2x\left(x+5\right)\left(x+5\right)^2}{x.2x\left(x+5\right)}\)
= \(\dfrac{3\left(x+5\right)^2}{x}\)
Bài 2:
c) \(\dfrac{9x^2y}{-15x^3y}\)
= -\(\dfrac{3.3x^2y}{5.3x^2yx}\)
= -\(\dfrac{3}{5x}\)
d) \(\dfrac{-15x\left(x-y\right)^2}{6x^2\left(x-y\right)^3}\)
= \(\dfrac{5.3x\left(x-y\right)^2}{2x.3x\left(x-y\right)^2\left(x-y\right)}\)
= \(\dfrac{5}{2x\left(x-y\right)}\)
= \(\dfrac{5}{2x^2-2xy}\)
Chứng minh đẳng thức:
\(\dfrac{a^3-4a^2-a+4}{a^3-7a^2+14a-8}=\dfrac{a+1}{a-2}\)
\(\dfrac{x^2y^2+1+\left(x^2-y\right)\left(1-y\right)}{x^2y^2+1+\left(x^2+y\right)\left(1+y\right)}=\dfrac{y^2-y+1}{y^2+y+1}\)
Lời giải:
1.
\(\frac{a^3-4a^2-a+4}{a^3-7a^2+14a-8}=\frac{a^2(a-4)-(a-4)}{(a^3-8)-(7a^2-14a)}=\frac{(a-4)(a^2-1)}{(a-2)(a^2+2a+4)-7a(a-2)}\)
\(=\frac{(a-4)(a-1)(a+1)}{(a-2)(a^2-5a+4)}=\frac{(a-4)(a-1)(a+1)}{(a-2)(a-1)(a-4)}=\frac{a+1}{a-2}\)
2.
\(\frac{x^2y^2+1+(x^2-y)(1-y)}{x^2y^2+1+(x^2+y)(1+y)}=\frac{x^2y^2+1+x^2-x^2y-y+y^2}{x^2y^2+1+x^2+x^2y+y+y^2}\)
\(=\frac{(x^2y^2-x^2y+x^2)+(y^2-y+1)}{(x^2y^2+x^2y+x^2)+(y^2+y+1)}\)
\(=\frac{x^2(y^2-y+1)+(y^2-y+1)}{x^2(y^2+y+1)+(y^2+y+1)}=\frac{(x^2+1)(y^2-y+1)}{(x^2+1)(y^2+y+1)}=\frac{y^2-y+1}{y^2+y+1}\)
quy đồng mẫu thức của các phân tử
a, \(\dfrac{2}{5x^3y};\dfrac{5}{xy^2}\)
b, \(\dfrac{3x}{x-5};\dfrac{-2}{3\left(x-5\right)}\)
c, \(\dfrac{1}{\left(x+3\right)\left(x-3\right)};\dfrac{5x}{x^2-9}\)
Chứng minh các bất đẳng thức sau với x, y, z > 0
a) \(x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2}\)
b) \(x^3+y^3\ge\dfrac{\left(x+y\right)^3}{4}\)
c) \(x^4+y^4\ge\dfrac{\left(x+y\right)^4}{8}\)
e) \(x^2+y^2+z^2\ge\dfrac{\left(x+y+z\right)^2}{3}\)
f) \(x^3+y^3+z^3\ge3xyz\)
a) \(x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2}\)
\(\Leftrightarrow2x^2+2y^2\ge\left(x+y\right)^2\Leftrightarrow x^2+y^2\ge2xy\)
\(\Leftrightarrow x^2-2xy+y^2\ge0\Leftrightarrow\left(x-y\right)^2\ge0\left(đúng\right)\)
b) \(x^3+y^3\ge\dfrac{\left(x+y\right)^3}{4}\)
\(\Leftrightarrow4x^3+4y^3\ge\left(x+y\right)^3\Leftrightarrow3x^3+3y^3\ge3x^2y+3xy^2\)
\(\Leftrightarrow3x^2\left(x-y\right)-3y^2\left(x-y\right)\ge0\)
\(\Leftrightarrow3\left(x-y\right)\left(x^2-y^2\right)\ge0\Leftrightarrow3\left(x-y\right)^2\left(x+y\right)\ge0\left(đúng\right)\)
a: Ta có: \(x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2}\)
\(\Leftrightarrow2x^2+2y^2-x^2-2xy-y^2\ge0\)
\(\Leftrightarrow x^2-2xy+y^2\ge0\)
\(\Leftrightarrow\left(x-y\right)^2\ge0\)(luôn đúng)
