Bạn chú thích hơi quá lố :) 

Ta có :( 5x - 3y + 4z ) . ( 5x - 3y - 4z ) \(=\left(5x-3y\right)^2-16z^2\)

\(=25x^2-30xy+9y^2-16z^2\)

Mà x^2=y^2 + z^2 nên ( 5x - 3y + 4z ) . ( 5x - 3y - 4z )\(=25x^2-30xy+9y^2-16\left(x^2-y^2\right)\)

\(=9x^2-30xy+25y^2=\left(3x-5y\right)^2\)

Học tốt !

\(\left(5x-3y+4z\right)\left(5x-3y-4z\right)=\left(3x-5y^2\right)\)

\(\Leftrightarrow\left(5x-3y\right)^2-16z^2-\left(3x-5y\right)^2=0\)

\(\Leftrightarrow\left(5x-3y-3x+5y\right)\left(5x-3y+3x-5y\right)-16z^2=0\)

\(\Leftrightarrow16x^2=16y^2+16z^2\)(luôn đúng)

 \(a^2-3ab+2c=0\)

\(\Leftrightarrow\left(x+y\right)^2-3\left(x+y\right)\left(x^2+y^2\right)+2\left(x^3+y^3\right)=0\)

\(\Leftrightarrow x^2+y^2+2xy-3x^3-3xy^2-3yx^2-3y^3+2y^3+2x^3=0\)

\(\Leftrightarrow\left(x+y\right)^2-\left(x+y\right)^3=0\)

Hình như đề hơi sai nếu là a^3 thì ms đúng chứ

mk chỉ góp ý thui tại làm mãi ko ra

hhihihihi

a) 128 - 3(x + 4) = 23

             3(x + 4) = 128 -23

             3(x + 4) = 105

                x + 4  = 105 : 3

                x + 4  = 35

                x        = 35 - 4

                x        = 31

\(a.128-3\left(x+4\right)=23.\)

\(3\left(x+4\right)=128-23\)

\(3\left(x+4\right)=105\)

\(x+4=105:3=35\)

\(x=35-4=31\)

\(b.\left[\left(4x+28\right)\cdot3+55\right]:5=35\)

\(\left(4x+28\right)\cdot3+55=35\cdot5=175\)

\(\left(4x+28\right)\cdot3=175-55=120\)

\(4x+28=120:3=40\)

\(4x=40-28=12\)

\(x=12:4=3\)

a, \(128-3.\left(x+4\right)=23\)

\(\Rightarrow\) \(3.\left(x+4\right)=105\)

\(\Rightarrow\) \(x+4=35\)

\(\Rightarrow\) \(x=31\)

b, \(\left[\left(4x+28\right).3+55\right]:5=35\)

\(\Rightarrow\) \(\left(4x+28\right).3+55=175\)

\(\Rightarrow\) \(\left(4x+28\right).3=120\)

\(\Rightarrow\) \(4x+28=40\)

\(\Rightarrow\) \(4x=12\)

\(\Rightarrow\) \(x=3\)

a: \(P=2x^2+3xy+y^2=\left(2x+y\right)\left(x+y\right)\)

\(=\left(2\cdot\dfrac{-1}{2}+\dfrac{2}{3}\right)\left(\dfrac{-1}{2}+\dfrac{2}{3}\right)\)

\(=\dfrac{-1}{3}\cdot\dfrac{1}{6}=-\dfrac{1}{18}\)

d: \(Q=\dfrac{-1}{3}x^4y^2=\dfrac{-1}{3}\cdot16\cdot\dfrac{1}{16}=-\dfrac{1}{3}\)

dài thế

dai the giai bao gio xong chac mua quyt nam sau moi giai xong

\(2^x+3\cdot2^x=9\cdot2^9\)

\(\Rightarrow2^x\cdot\left(1+3\right)=9\cdot2^9\)

\(\Rightarrow2^x\cdot4=9\cdot2^9\)

\(\Rightarrow2^x=\dfrac{9\cdot2^9}{4}\)

\(\Rightarrow2^x=9\cdot2^7\)

Xem lại đề ! 

f) = x²( x - 4 ) - 9( x - 4 ) = ( x - 4 )( x - 3 )( x + 3 )

g) = 4( x - y ) + ( x - y )² = ( x - y )( x - y + 4 )

h) = x³( x + 1 ) + ( x - 1 )( x + 1 ) = ( x + 1 )( x³ + x - 1 )

i) = ( x - y )( x + y ) - 4( x + y ) = ( x + y )( x - y - 4 )

j) = ( x - y )( x² + xy + y² ) - 3( x - y ) = ( x - y )( x² + xy + y² - 3 )

Trả lời:

f, x³ - 4x² - 9x + 36 = ( x³ - 4x² ) - ( 9x - 36 ) = x² ( x - 4 ) - 9 ( x - 4 ) = ( x - 4 )( x² - 9 ) = ( x - 4 )( x - 3 )( x + 3 )

g, 4x - 4y + x² - 2xy + y² = ( 4x - 4y ) + ( x² - 2xy + y² ) = 4 ( x - y ) + ( x - y )² = ( x - y ) ( 4 + x - y )

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

i, x² - y² - 4x - 4y = ( x² - y² ) - ( 4x + 4y ) = ( x - y )( x + y ) - 4 ( x + y ) = ( x + y )( x - y - 4 )

j, x³ - y³ - 3x + 3y = ( x³ - y³ ) - ( 3x - 3y ) = ( x - y )( x² + xy + y² ) - 3 ( x - y ) = ( x - y )( x² + xy + y² - 3 ) 

f) x³-4x²-9x+36

=x²(x-4)-9(x-4)

=(x-4)(x²-9)

=(x-4)(x-3)(x+3)

g) 4x-4y+x²-2xy+y²

=4(x-y)+(x-y)²

=(x-y)(4+x-y)

h) x⁴+x³+x²-1

=x³(x+1)+(x-1)(x+1)

=(x+1)(x³+x-1)

i) x²-y²-4x-4y

=(x-y)(x+y)-4(x+y)

=(x+y)(x-y-4)

j) x³-y³-3x+3y

=(x-y)(x²+xy+y²)-3(x-y)

=(x-y)(x²+xy+y²-3)

#H

ở bài 1 đầu bài là viết các tich và các thương sau dưới dạng lũy thừa mình viết thiếu 

bạn đăng từng bài một thôi nhé

Bài 1:

a) 8² . 2² = 64 . 4 = 256

b) 2⁵. 4 = 2⁵. 2²= 2⁵⁺² = 2⁷

c) 9⁴. 243 = 9⁴. 3⁵=(3²)⁴.3⁵=3⁸.3⁵=3⁸⁺⁵=3¹³

Bài 1 :

\(a.8^2.2^2=\left(8.2\right)^2=16^2\)

\(b.2^5.4=2^5.2^2=2^7\)

\(c.9^4.243=\left(3^2\right)^4.3^5=3^8.3^5=3^{13}\)

Bài 2:

a. ( đề sai =.= )

\(b.12.\left(x-1\right):3=43+2^3\)

\(\Leftrightarrow12.\left(x-1\right)=43+2^3.3\)

\(\Leftrightarrow12.\left(x-1\right)=67\)

\(\Leftrightarrow x-1=\frac{67}{12}\)

\(\Leftrightarrow x=\frac{79}{12}\)

\(x^3-3x^2+3x-y^3-1\)

\(=\left(x-1\right)^3-y^3\)

\(=\left(x-1-y\right).\left[\left(x-1\right)^2+\left(x-1\right).y+y^2\right]\)