cửa hàng đó có 0,9 tấn kg gạo đúng ko

Sửa đề : ... tấn gạo tẻ ....

7/10 tấn = 700kg; 1/5 tấn = 200kg

Cửa hàng đó có số gạo là : 700 + 200 = 900 ( kg gạo )

Đ/s : 900kg gạo
 

a) \(a^3+b^3-c^3+3abc\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)-c^3+3abc\)

\(=\left(a+b-c\right)\left[\left(a+b\right)^2+\left(a+b\right)c+c^2\right]-3ab\left(a+b-c\right)\)

\(=\left(a+b-c\right)\left(a^2+2ab+b^2+ca+bc+c^2-3ab\right)\)

\(=\left(a+b-c\right)\left(a^2+b^2+c^2-ab+bc+ca\right)\)

b) \(\left(x+y+z\right)^3-x^3-y^3-z^3\)

\(=\left(x+y+z-x\right)\left[\left(x+y+z\right)^2+\left(x+y+z\right)x+x^2\right]-\left(y+z\right)\left(y^2-yz+z^2\right)\)

\(=\left(y+z\right)\left[x^2+y^2+z^2+2\left(xy+yz+zx\right)+x^2+xy+zx+x^2-y^2+yz-z^2\right]\)

\(=\left(y+z\right)\left(3x^2+3xy+3yz+3zx\right)\)

\(=3\left(y+z\right)\left[x\left(x+y\right)++z\left(x+y\right)\right]\)

\(=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

\(a)a^3+b^3-c^3+3abc=\left(a+b\right)^3-3ab\left(a+b\right)-c^3+3abc\)

\(=\left(a+b-c\right)\left[\left(a+b\right)^2+\left(a+b\right)c+c^2\right]-3ab\left(a+b-c\right)\)

\(=\left(a+b-c\right)\left(a^2+2ab+b^2+ac+bc+c^2-3ab\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab+bc+ac\right)\)

\(\frac{7}{8}+x=\frac{3}{5}\Rightarrow x=\frac{3}{5}-\frac{7}{8}\Rightarrow x=\frac{-11}{40}\)

Vậy x= -11/40

\(\left|x+1\right|=2\Rightarrow\orbr{\begin{cases}x+1=2\\x+1=-2\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}}\)

Vậy x=1 hoặc x=-3

\(a,\frac{7}{8}+x=\frac{3}{5}\Rightarrow x=\frac{3}{5}-\frac{7}{8}\Rightarrow x=-\frac{1}{4}\)

\(|x+1|=2\Rightarrow x+1=2\Rightarrow x=2-1\Rightarrow x=1\)

a) \(4x^3-9x=0\)

\(\Leftrightarrow x\left(4x^2-9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\4x^2=9\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm\frac{3}{2}\end{cases}}\)

b) \(3x\left(x-2\right)-5x+10=0\)

\(\Leftrightarrow\left(3x-5\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=2\end{cases}}\)

c) \(4x\left(x+3\right)-x^2+9=0\)

\(\Leftrightarrow4x\left(x+3\right)-\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left(3x+3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-3\end{cases}}\)

d) \(\left(2x+5\right)\left(x-4\right)=\left(x-4\right)\left(5-x\right)\)

\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)+\left(x-5\right)\left(x-4\right)=0\)

\(\Leftrightarrow3x\left(x-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)

e) \(16x^2-25=\left(4x-5\right)\left(2x+1\right)\)

\(\Leftrightarrow\left(4x-5\right)\left(4x+5\right)-\left(4x-5\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left(4x-5\right)\left(2x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{4}\\x=-2\end{cases}}\)

f) \(\left(x+\frac{1}{5}\right)^2=\frac{64}{9}\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{5}=\frac{8}{3}\\x+\frac{1}{5}=-\frac{8}{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{37}{15}\\x=-\frac{43}{15}\end{cases}}\)

g) \(9\left(x+2\right)^2=\left(x+3\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}3x+6=x+3\\3x+6=-x-3\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=-3\\4x=-9\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=-\frac{9}{4}\end{cases}}\)

a) 4x³ - 9x = 0

<=> x( 4x² - 9 ) = 0

<=> x( 2x - 3 )( 2x + 3 ) = 0

<=> x = 0 hoặc 2x - 3 = 0 hoặc 2x + 3 = 0

<=> x = 0 hoặc x = ±3/2

b) 3x( x - 2 ) - 5x + 10 = 0

<=> 3x( x - 2 ) - 5( x - 2 ) = 0

<=> ( x - 2 )( 3x - 5 ) = 0

<=> x - 2 = 0 hoặc 3x - 5 = 0

<=> x = 2 hoặc x = 5/3

c) 4x( x + 3 ) - x² + 9 = 0

<=> 4x( x + 3 ) - ( x² - 9 ) = 0

<=> 4x( x + 3 ) - ( x - 3 )( x + 3 ) = 0

<=> ( x + 3 )[ 4x - ( x - 3 ) ] = 0

<=> ( x + 3 )( 4x - x + 3 ) = 0

<=> ( x + 3 )( 3x + 3 ) = 0

<=> x + 3 = 0 hoặc 3x + 3 = 0

<=> x = -3 hoặc x= -1

d) ( 2x + 5 )( x - 4 ) = ( x - 4 )( 5 - x )

<=> ( 2x + 5 )( x - 4 ) - ( x - 4 )( 5 - x ) = 0

<=> ( x - 4 )[ ( 2x + 5 ) - ( 5 - x ) ] = 0

<=> ( x - 4 )( 2x + 5 - 5 + x ) = 0

<=> ( x - 4 ).3x = 0

<=> x - 4 = 0 hoặc 3x = 0

<=> x = 4 hoặc x = 0

e) 16x² - 25 = ( 4x - 5 )( 2x + 1 )

<=> ( 4x - 5 )( 4x + 5 ) - ( 4x - 5 )( 2x + 1 ) = 0

<=> ( 4x - 5 )[ ( 4x + 5 ) - ( 2x + 1 ) ] = 0

<=> ( 4x - 5 )( 4x + 5 - 2x - 1 ) = 0

<=> ( 4x - 5 )( 2x + 4 ) = 0

<=> 4x - 5 = 0 hoặc 2x + 4 = 0

<=> x = 5/4 hoặc x = -2

f) ( x + 1/5 )² = 64/9

<=> ( x + 1/5 )² = ( ±8/3 )²

<=> x + 1/5 = 8/3 hoặc x + 1/5 = -8/3

<=> x = 37/15 hoặc x = -43/15

g) 9( x + 2 )² = ( x + 3 )²

<=> 3²( x + 2 )² - ( x + 3 )² = 0

<=> [ 3( x + 2 ) ]² - ( x + 3 )² = 0

<=> ( 3x + 6 )² - ( x + 3 )² = 0

<=> [ ( 3x + 6 ) - ( x + 3 ) ][ ( 3x + 6 ) + ( x + 3 ) ] = 0

<=> ( 3x + 6 - x - 3 )( 3x + 6 + x + 3 ) = 0

<=> ( 2x + 3 )( 4x + 9 ) = 0

<=> 2x + 3 = 0 hoặc 4x + 9 = 0

<=> x = -3/2 hoặc x = -9/4

a) p² - 2pq + q² - 9 ( sửa rồi nhé :)) )

= ( p² - 2pq + q² ) - 9

= ( p - q )² - 3²

= ( p - q - 3 )( p - q + 3 )

b) ( a + 3 )² + ( a² - 9 )² ( sửa rồi nhé p2 :)) )

= ( a + 3 )² + [ ( a - 3 )( a + 3 ) ]²

= ( a + 3 )² + ( a - 3 )²( a + 3 )²

= ( a + 3 )²[ 1 + ( a - 3 )² ]

= ( a + 3 )²( 1 + a² - 6a + 9 )

= ( a + 3 )²( a² - 6a + 10 )

(2-1/2):(-3/4+1/2)

=(4/2-1/2):(-3/4+2/4)

=3/2:(-1/4)

=(-6)

\(\frac{3}{2}:(-\frac{1}{4})=\frac{-12}{2}=\frac{-6}{1}\)