a: \(=\dfrac{7}{5}\cdot\dfrac{15}{49}-\dfrac{12+10}{15}:\dfrac{11}{5}\)

\(=\dfrac{3}{7}-\dfrac{22}{15}\cdot\dfrac{5}{11}=\dfrac{3}{7}-\dfrac{2}{3}=\dfrac{9-14}{21}=\dfrac{-5}{21}\)

b: =>2,8x-32=-60

=>2,8x=-28

hay x=-10

4× (x+3)= 2 ×(x+1)+20

4 x + 12 = 2x + 2 + 20

2x = 10

x = 5

4.(x+3)=2.x+2+20

12+4x =2x+22

4x       =2x+10

2x       =10

x         =10:2

x         =5

Nhớ phải giữ lời hứa đó!

Học tốt!

\(\frac{x-2}{16}=\frac{-4}{2-x}\)

\(\Leftrightarrow\frac{x-2}{16}+\frac{4}{2-x}=0\)

\(\Leftrightarrow\frac{\left(x-2\right)\left(2-x\right)+4.16}{16\left(2-x\right)}=0\)

\(\Leftrightarrow\left(x-2\right)\left(2-x\right)+64=0\)

\(\Leftrightarrow-\left(x-2\right)\left(x-2\right)=-64\)

\(\Leftrightarrow-\left(x-2\right)^2=-8^2\)

\(\Leftrightarrow\left(x-2\right)^2=8^2\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=8\\x-2=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}x=10\\x=-6\end{cases}}}\)

P/s: Mik nghĩ bài này lớp 8 thì đúng hơn vì nó liên quan đến hằng đẳng thức

Nếu là lp 8 thì giải theo cách này nha:

\(\Rightarrow\left(x-2\right).\left(2-x\right)=16.\left(-4\right)\)

\(2x-x^2-4+2x=-64\)

\(-x^2+4x-4=-64\)

\(-\left(x+2\right)^2=-64\)

\(\Rightarrow\left(x+2\right)^2=8^2\)

\(\Rightarrow\orbr{\begin{cases}x+2=8\\x+2=-8\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=6\\x=-10\end{cases}}\)

Nâng cao lớp 7 cũng là chuyên đề Hằng đẳng thức mà bạn Moon!!

a)\(-3x\left(x+2\right)^2+\left(x+3\right)\left(x-1\right)\left(x+1\right)-\left(2x-3\right)^2\)

\(=-3x.\left(x^2+2.x.2+2^2\right)+\left(x^2+x+3x-3\right).\left(x+1\right)-\left(2x\right)^2-2.2.x.\left(-3\right)+\left(-3\right)^2\)

\(=-3x.\left(x^2+4x+4\right)+\left(x^2+\left(x+3x\right)-3\right).\left(x+1\right)-4x+12x+9\)

\(=-3x.\left(x^2+4x+4\right)+\left(x^2+4x-3\right)\left(x+1\right)-4x+12x+9\)

\(=-3x^3-12x^2-12x+x^3+4x^2-3x+x^2+4x-3-4x+12x+9\)

\(=\left(-3x^3-x^3\right)+\left(-12x^2+4x^2+x^2\right)+\left(-12x-3x+4x-4x+12x\right)+\left(-3+9\right)\)

\(=-2x^3-7x^2-3x+6\)

b)\(\left(x-3\right)\left(x+3\right)\left(x+2\right)-\left(x-1\right)\left(x^2-3\right)-5x\left(x+4\right)^2-\left(x-5\right)^2\)

\(=\left(x.\left(x+3\right)-3\left(x+3\right)\right)\left(x+2\right)-\left(x.\left(x^2-3\right)-1\left(x^2-3\right)\right)-5x\left(x+4\right)^2-\left(x-5\right)^2\)

\(=\left(x.x+x.3-3.x+\left(-3\right).3\right)\left(x+2\right)-\left(x.x^2+x.\left(-3\right)-1.x^2+\left(-1\right).\left(-3\right)\right)-5x.x+\left(-5x\right).4-x^2-2x5+5^2\)

\(=\left(x^2+3x-3x-9\right)\left(x+2\right)-x^3-3x-x^2+3-5x^2-20x-x^2-10x+25\)

\(=\left(x^2+\left(3x-3x\right)-9\right)\left(x+2\right)-x^3-3x-x^2+3-5x^2-20x-x^2-10x+25\)

\(=\left(x^2-9\right)\left(x+2\right)-x^3-3x-x^2+3-5x^2-20x-x^2-10x+25\)

\(=x^3+2x^2-9x-15-x^3-3x-x^2+3-5x^2-20x-x^2-10x+25\)

\(=\left(x^3-x^3\right)+\left(2x^2-x^2-5x^2-x^2\right)+\left(-9x-3x-20x-10x\right)+\left(-18+3+25\right)\)

\(=-5x^2-42x+10\)

dễ

hết

chỗ

chê

\(1-\)\(a,\frac{7}{8}-\frac{2}{3}=\frac{21}{24}-\frac{16}{24}=\frac{5}{24}\)

\(b,\frac{7}{8}-\frac{2}{3}\times\frac{3}{4}=\frac{7}{8}-\frac{1}{2}=\frac{7}{8}-\frac{4}{8}=\frac{3}{8}\)

\(2-\)\(a,\frac{5}{8}-x=\frac{5}{9}\)

\(x=\frac{5}{8}-\frac{5}{9}\)

\(x=\frac{5}{72}\)

\(b,\frac{3}{4}:x=\frac{2}{5}\)

\(x=\frac{3}{4}:\frac{2}{5}\)

\(x=\frac{15}{8}\)

1, Tính:

\(\frac{7}{8}\) - \(\frac{2}{3}\) = \(\frac{21}{24}\) - \(\frac{16}{24}\) = \(\frac{5}{24}\)

\(\frac{7}{8}\) - \(\frac{2}{3}\) x \(\frac{3}{4}\) = \(\frac{7}{8}\) - \(\frac{1}{2}\) = \(\frac{7}{8}\) - \(\frac{4}{8}\) = \(\frac{3}{8}\)

2, Tìm x

\(\frac{5}{8}\)- x = \(\frac{5}{9}\)

           x = \(\frac{5}{8}\)- \(\frac{5}{9}\)

           x = \(\frac{5}{72}\)

\(\frac{3}{4}\): x = \(\frac{2}{5}\)

           x = \(\frac{3}{4}\) : \(\frac{2}{5}\)

           x = \(\frac{15}{8}\)

\(\left(8x-4x^2-1\right)\left(x^2+2x-1\right)=4\left(x^2+x+1\right)\)

\(11x^2+6x-4x^4-1=4x^2+4x+4\)

\(11x^2+6x-4x^4-4x^2-4x-4=0\)

\(7x^2+2x-4x^4-5=0\)

\(\left(x-1\right)\left(x-1\right)\left(-4x^2-8x-5\right)=0\)

bn lm nốt nha , ko có dấu hoặc nên mk làm đến đây thôi 

Cảm ơn nha ! Nhưng sao mình ko ấn đúng cho bạn được !? hic