xin lỗi tớ ấn nhầm chỗ M=7 tớ làm lại rồi đó 

ban tra loi het cac cau hoi phia tren kia ho minh dc ko?
 

a) ( 2x + 3 )² - 2( 2x + 3 )( 2x + 5 ) + ( 2x + 5 )²

= [ ( 2x + 3 ) - ( 2x + 5 ) ]²

= ( 2x + 3 - 2x - 5 )²

= (-2)² = 4

b) ( x² + x + 1 )( x² - x + 1 )( x² - 1 )

= ( x⁴ - x³ + x² + x³ - x² + x + x² - x + 1 )( x² - 1 )

= ( x⁴ + x² + 1 )( x² - 1 )

= x⁶ - x⁴ + x⁴ - x² + x² - 1

= x⁶ - 1 

c) ( x + y )² + ( x - y )²

= x² + 2xy + y² + x² - 2xy + y²

= 2x² + 2y² = 2( x² + y² )

d) 2( x - y )( x + y ) + ( x + y )² + ( x - y )²

= [ ( x + y ) + ( x - y ) ]²

= ( x + y + x - y )²

= ( 2x )² = 4x²

e) ( x - y + z )² + ( z - y )² + 2( x - y + z )( y - z )

= ( x - y + z )² + ( z - y )² - 2( x - y + z )( z - y )

= [ ( x - y + z ) - ( z - y ) ]²

= ( x - y + z - z + y )²

= x²

f) ( a + b - c )² + ( a - b + c )² - 2( b - c )²

= [ ( a + b ) - c ]² + [ ( a - b ) + c ]² - 2( b² - 2bc + c² )

= [ ( a + b )² - 2( a + b )c + c² ] + [ ( a - b )² + 2( a - b )c + c² ] - 2b² + 4bc - 2c²

= a² + b² + c² + 2ab - 2bc - 2ca + c² + a² + b² + c² - 2ab + 2bc + 2ac - 2b² + 4bc - 2c²

= 2a² 

g) ( a + b + c )² + ( a - b - c )² + ( b - c - a )² + ( c - a - b )²

= [ ( a + b ) + c ]² + [ ( a - b ) - c ]² + [ ( b - c ) - a ]² + [ ( c - a ) - b ]²

= [ ( a + b )² + 2( a + b )c + c² ] + [ ( a - b )² - 2( a - b )c + c² ] + [ ( b - c )² - 2( b - c )a + a² ] + [ ( c - a )² - 2( c - a )b + b² ]

= [ a² + b² + c² + 2ab + 2bc + 2ca ] + [ a² + b² + c² - 2ab + 2bc - 2ca ] + [ a² + b² + c² - 2ab - 2bc + 2ca ] + [ a² + b² + c² + 2ab - 2bc - 2ca ]

= 4a² + 4b² + 4c²

Có vẻ hơi dài dòng nhỉ :( Nhưng như này là kĩ nhất đấy :)

Cảm ơn bạn Trần Nhật Quỳnh nhiều nhé:)

2: Ta có: \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=\dfrac{a\left(a+b+c\right)}{b+c}+\dfrac{b\left(a+b+c\right)}{c+a}+\dfrac{c\left(a+b+c\right)}{a+b}-a-b-c=\left(a+b+c\right)\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)=a+b+c-a-b-c=0\)

1: Sửa đề: Cho \(x,y,z\ne0\) và \(\dfrac{1}{x}+\dfrac{2}{y}+\dfrac{1}{z}=\dfrac{2}{2x+y+2z}\).

CM:....

Đặt 2x = x', 2z = z'.

Ta có: \(\dfrac{2}{x'}+\dfrac{2}{y}+\dfrac{2}{z'}=\dfrac{2}{x'+y+z'}\)

\(\Leftrightarrow\dfrac{1}{x'}+\dfrac{1}{y}+\dfrac{1}{z'}=\dfrac{1}{x'+y+z'}\)

\(\Leftrightarrow\dfrac{1}{x'}-\dfrac{1}{x'+y+z'}+\dfrac{1}{y}+\dfrac{1}{z'}=0\)

\(\Leftrightarrow\dfrac{y+z'}{x'\left(x'+y+z'\right)}+\dfrac{y+z'}{yz'}=0\)

\(\Leftrightarrow\dfrac{\left(y+z'\right)\left(yz'+x'^2+x'y+x'z'\right)}{x'yz'\left(x'+y+z'\right)}=0\)

\(\Leftrightarrow\dfrac{\left(x'+y\right)\left(y+z'\right)\left(z'+x'\right)}{x'yz'\left(x'+y+z'\right)}=0\Leftrightarrow\left(2x+y\right)\left(y+2z\right)\left(2z+2x\right)=0\Leftrightarrow\left(2x+y\right)\left(y+2z\right)\left(z+x\right)=0\left(đpcm\right)\)

ChươngII *Dạng toán rútg gọn phân thức

Bài 1.Rút gọn phân thức

a. \(\dfrac{3x\left(1-x\right)}{2\left(x-1\right)}=\dfrac{-3x\left(x-1\right)}{2\left(x-1\right)}=-\dfrac{3x}{2}\)

b.\(\dfrac{6x^2y^2}{8xy^5}=\dfrac{3x.2xy^2}{4y^3.2xy^2}=\dfrac{3x}{4y^3}\)

c.\(\dfrac{23\left(x-y\right)\left(x-z\right)^2}{6\left(x-y\right)\left(x-z\right)}=\dfrac{23\left(x-z\right)}{6}\)

Bài 2 rút gọn các phân thức sau:

a.\(\dfrac{x^2-16}{4x-x^2}=\dfrac{\left(x-4\right)\left(x+4\right)}{-x\left(x-4\right)}=-\dfrac{x+4}{x}\)(x khác 0,x khác 4)

b.\(\dfrac{x^2+4x+3}{2x+6}=\dfrac{x^2+3x+x+3}{2\left(x+3\right)}=\dfrac{\left(x+3\right)\left(x+1\right)}{2\left(x+3\right)}=\dfrac{x+1}{2}\)

( x \(\ne-3\) )

c.\(\dfrac{15x\left(x+y\right)^3}{5y\left(x+y\right)^2}=\dfrac{3x\left(x+y\right)}{y}\) (y+(x+y) khác 0)

d. \(\dfrac{5\left(x-y\right)-3\left(y-x\right)}{10\left(x-y\right)}=\dfrac{5\left(x-y\right)+3\left(x-y\right)}{10\left(x-y\right)}=\dfrac{8\left(x-y\right)}{10\left(x-y\right)}=\dfrac{4}{5}\)

(x khác y)

e.\(\dfrac{2x+2y+5x+5y}{2x+2y-5x-5y}=\dfrac{2\left(x+y\right)+5\left(x+y\right)}{2\left(x+y\right)-5\left(x+y\right)}=\dfrac{7\left(x+y\right)}{-3\left(x+y\right)}=-\dfrac{7}{3}\)

(x khác -y)

f.\(\dfrac{x^2-xy}{3xy-3y^2}=\dfrac{x\left(x-y\right)}{3y\left(x-y\right)}=\dfrac{x}{3y}\)(x khác y,y khác 0)

g.\(\dfrac{2ax^2-4ax+2a}{5b-5bx^2}=\dfrac{2a\left(x^2-2x+1\right)}{-5b\left(x^2-1\right)}=\dfrac{2a\left(x-1\right)^2}{-5b\left(x-1\right)\left(x+1\right)}=\dfrac{2a\left(x-1\right)}{-5b\left(x+1\right)}\)

\ (b khác 0,x khác +-1)

h. \(\dfrac{4x^2-4xy}{5x^3-5x^2y}=\dfrac{4x\left(x-y\right)}{5x^2\left(x-y\right)}=\dfrac{4x}{5x^2}\)

(x khác 0,x khác y)

i.\(\dfrac{\left(x+y\right)^2-z^2}{x+y+z}=\dfrac{\left(x+y+z\right)\left(x+y-z\right)}{x+y+z}=x+y-z\)

(x+y+z khác 0)

k.\(\dfrac{x^6+2x^3y^3+y^6}{x^7-xy^6}=\dfrac{\left(x^3\right)^2+2x^3y^3+\left(y^3\right)^2}{x\left(x^6-y^6\right)}=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^3-y^3\right)\left(x^3+y^3\right)}=\dfrac{x^3+y^3}{x\left(x^3-y^3\right)}\)

(x khác 0,x khác +-y)

Bài 4 : Rút gọn các phân thức sau :

\(a,\dfrac{\left(a+b\right)^2-c^2}{a+b+c}=\dfrac{\left(a+b+c\right)\left(a+b-c\right)}{a+b+c}=a+b-c\)

\(b,\dfrac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}=\dfrac{\left(a^2+2ab+b^2\right)-c^2}{\left(a^2+2ac+c^2\right)-b^2}\)

\(=\dfrac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\dfrac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(a+c-b\right)}=\dfrac{a+b-c}{a+c-b}\)

c,\(\dfrac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}\)

\(=\dfrac{\left(2x^3-x^2-15x\right)-\left(6x^2-3x-45\right)}{\left(3x^3-10x^2+3x\right)-\left(9x^2-30x+9\right)}\)

\(=\dfrac{x\left(2x^2-x-15\right)-3\left(2x^2-x-15\right)}{x\left(3x^2-10x+3\right)-3\left(3x^2-10x+3\right)}\)
\(=\dfrac{\left(x-3\right)\left(2x^2-x-15\right)}{\left(x-3\right)\left(3x^2-10x+3\right)}\)

\(=\dfrac{\left(x-3\right)\left(2x^2+5x-6x-15\right)}{\left(x-3\right)\left(3x^2-9x-x+3\right)}\)
\(=\dfrac{\left(x-3\right)\left[x\left(2x+5\right)-3\left(2x+5\right)\right]}{\left(x-3\right)\left[3x\left(x-3\right)-\left(x-3\right)\right]}\)

\(=\dfrac{\left(x-3\right)^2\left(2x+5\right)}{\left(x-3\right)^2\left(3x-1\right)}\)

\(=\dfrac{2x+5}{3x-1}\)

a, = -3/2

b, = x-z/2

c, = (x-4).(x+4)/-x.(x-4) = -(x+4)/x = -x-4/x

k mk nha

Câu a.

Ta luôn có 

\(\frac{a}{a+b}>\frac{a}{a+b+c}\)  (do a+b < a+b+c)

\(\frac{b}{b+c}>\frac{b}{a+b+c}\)

\(\frac{c}{c+a}>\frac{c}{a+b+c}\)

Cộng theo từng vế rồi rút gọn ta đươc đpcm

Cảm ơn b nhé. B biết làm.câu b c d không giúp m với

b/  \(\left(x+y+z\right)\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\)

\(=\frac{1}{2}.\left(\left(x+y\right)+\left(y+z\right)+\left(z+x\right)\right)\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\)

\(\frac{1}{2}.3\sqrt[3]{\left(x+y\right)\left(y+z\right)\left(z+x\right)}.\frac{3}{\sqrt[3]{\left(x+y\right)\left(y+z\right)\left(z+x\right)}}=\frac{9}{2}\)