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Bài 1:
a) \(\frac{4}{9}x^2-y^2=\left(\frac{2}{3}x-y\right)\left(\frac{2}{3}x+y\right)\)
b) \(x^2-5=\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)\)
c) \(4x^2+6x+9=\left(2x+2\right)^2+5\)ko hiểu ???
d) \(\frac{1}{9}x^2-\frac{4}{3}xy+4=\left(\frac{1}{3}x\right)^2-2.\frac{1}{3}x.2+2^2=\left(\frac{1}{3}x-2\right)^2\)
Bài 2:
a) \(\left(\frac{1}{2}x-\frac{1}{3}y\right)\left(\frac{1}{2}x+\frac{1}{3}y\right)=\frac{1}{4}x^2-\frac{1}{9}y^2\)
b) \(\left(2x-\frac{1}{3}y\right)\left(4x^2+\frac{2}{3}xy+\frac{1}{9}x^2\right)=8x^3-\frac{1}{27}y^3\)
c) \(\left(3x-5y\right)\left(9x^2+15xy+\frac{1}{9}x^2\right)=27x^3-125y^3\)
\(=2x^2y-\frac{1}{2}x^2y^2-xy\left(2x-xy\right)\)
\(=xy\left(2x-\frac{1}{2}xy\right)-xy\left(2x-xy\right)\)
\(=xy\left(2x-\frac{1}{2}xy-2x+xy\right)\)
\(=xy.\frac{1}{2}xy\)
\(=\frac{1}{2}x^2y^2\)
\(d,\frac{10x+3}{8}=\frac{7-8x}{12}\)
\(\left(10x+3\right):8=\left(7-8x\right):12\)
\(\left(10x+3\right).\frac{1}{8}=\left(7-8x\right).\frac{1}{12}\)
\(\frac{5}{4}x+\frac{3}{8}=\frac{7}{12}-\frac{8}{12}x\)
\(\frac{5}{4}x+\frac{8}{12}x=\frac{7}{12}-\frac{3}{8}\)
\(\frac{23}{12}x=\frac{5}{24}\)
\(x=\frac{5}{46}\)
E mới lớp 6 nên giải sai thì thông cảm ạ UwU
\(b,\frac{x}{10}-\left(\frac{x}{30}+\frac{2x}{45}\right)=\frac{4}{5}\)
\(< =>\frac{9x}{90}-\frac{7x}{90}=\frac{4}{5}\)
\(< =>\frac{x}{45}=\frac{32}{45}\)
\(< =>x=32\)
\(d,\frac{10x+3}{8}=\frac{7-8x}{12}\)
\(< =>\left(10x+3\right).12=\left(7-8x\right).8\)
\(< =>120x+36=56-64x\)
\(< =>184x=56-36=20\)
\(< =>x=\frac{20}{184}=\frac{5}{46}\)
\(a.\left(x^2+\frac{2}{5}y\right)\left(x^2-\frac{2}{5}y\right)\\ =x^4-\frac{4}{25}y^2\)
\(b.\left(2x+y^2\right)^3\\ =8x^3+12x^2y^2+6xy^4+y^6\)
\(c.\left(3x^2-2y\right)^3\\ =27x^6-54x^4y+36x^2y^2-8y^3\)
\(\left(x+4\right)\left(x^2-4x+16\right)\\ =x^3+64\)
\(e.\left(x^2-\frac{1}{3}\right)\left(x^4+\frac{1}{3}x^2+\frac{1}{9}\right)\\ =x^6-\frac{1}{27}\)
\(1,\left(\frac{a}{3}+4y\right)^2=\frac{a^2}{9}+\frac{8ay}{3}+16y^2\)
\(2,\)Bạn xem lại đề bài giùm mk nhé
\(\left(x^2+\frac{2}{5}y\right).\left(x^2-\frac{2}{5}y\right)=\left(x^2\right)^2-\left(\frac{2}{5}y\right)^2=x^4-\frac{4}{25}y^2\)