a: \(\sqrt{252}+\dfrac{1}{3}\sqrt{63}-\sqrt{175}\)

\(=4\sqrt{7}+\sqrt{7}-5\sqrt{7}\)

=0

a:

-2/5 + 4/5.x = 3/5

2/5 . x = 3/5

x = 3/5 : 2/5

x = 15/10

\(a,\dfrac{-2}{5}+\dfrac{4}{5}x=\dfrac{3}{5}\)

\(\Rightarrow\dfrac{4x-2}{5}=\dfrac{3}{5}\)

\(\Rightarrow4x-2=3\)

\(\Rightarrow4x=5\)

\(\Rightarrow x=\dfrac{5}{4}\)

\(b,\dfrac{-3}{7}-\dfrac{4}{7}:x=-2\)

\(\Rightarrow\dfrac{-4}{7x}=\dfrac{-11}{7}\)

\(\Rightarrow7x=\dfrac{-4.7}{-11}\)

\(\Rightarrow7x=\dfrac{28}{11}\)

\(\Rightarrow x=\dfrac{4}{11}\)

a,\(3xy+6x^2=3x\left(y+2x\right)\)

b,\(-4xy^2+8x^2y=-4xy\left(y-2x\right)\)

c,\(4\left(x+y\right)+x^2+xy=4\left(x+y\right)+x\left(x+y\right)\)\(=\left(4+x\right)\left(x+y\right)\)

d,\(4x^2-9y^2=\left(2x\right)^2-\left(3y\right)^2=\left(2x-3y\right)\left(2x+3y\right)\)

e,\(25x^2-\dfrac{1}{4}=\left(5x\right)^2-\left(\dfrac{1}{2}\right)^2=\left(5x-\dfrac{1}{2}\right)\left(5x+\dfrac{1}{2}\right)\)

a. \(=3x\left(y+2x\right)\)

b. \(=4xy\left(-y+2x\right)\)

c. \(=4\left(x+y\right)+x\left(x+y\right)=\left(4+x\right)\left(x+y\right)\)

d. \(=\left(2x+3y\right)\left(2x-3y\right)\)

e. \(=\left(5x+\dfrac{1}{2}\right)\left(5x-\dfrac{1}{2}\right)\)

f. \(=\left(4x+4+3x-6\right)\left(4x+4-3x+6\right)=\left(7x-2\right)\left(x+10\right)\)

g . \(=x^2+x+9x+9=x\left(x+1\right)+9\left(x+1\right)=\left(x+9\right)\left(x+1\right)\)

\(\dfrac{a}{b}=-\dfrac{4}{9}\)

11)11) 3x(x-5)²-(x+2)³+2(x-1)³-(2x+1)(4x²-2x+1)=3x(x²-10x+25)-(x³+6x²+12x+8)+2(x³-3x²+3x-1)-(8x³+1)=3x³-30x²+75x-x³-6x²-12x-8+2x³-6x²+6x-2-8x³-1=-4x³-42x²+63x-11

nhìn khó thế

mik điên vì nó lắm rùi

mk cx z nà -_-

bực lém

@Thúi Thị Thơm, công nhận cái tên của bạn độc đáo thiệt, mà bạn chặn quảng cáo đi, tớ hết rồi!!