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a)
a(a2+ab+b2)−b(a2+ab+b2)
a3+a2b+ab2−b(a2+ab+b2)
a3+a2b+ab2−(ba2+b2a+b3)
a3+a2b+ab2−ba2−b2a−b3
a3+a2b+ab2−ba2−b2a−b3
a3−b3
đợi làm b và c
b ) dễ hihi
8ac+2ad−12bc−3bd
c) chả hiểu đệ j cả
\(\frac{a}{2}\)=\(\frac{b}{\frac{3}{2}}\)=\(\frac{c}{\frac{4}{3}}\)=\(\frac{a-b}{2-\frac{3}{2}}\)=\(\frac{15}{\frac{1}{2}}\)=30
a=30.2=60
b=30.\(\frac{3}{2}\)=45
c=30.\(\frac{4}{3}\)=40
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>\(a=bk;c=dk\)
1: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2\cdot bk+3\cdot dk}{2b+3d}=\dfrac{k\left(2b+3d\right)}{2b+3d}=k\)
\(\dfrac{2a-3c}{2b-3d}=\dfrac{2bk-3dk}{2b-3d}=\dfrac{k\left(2b-3d\right)}{2b-3d}=k\)
Do đó: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)
2: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4\cdot bk-3b}{4\cdot dk-3d}=\dfrac{b\left(4k-3\right)}{d\left(4k-3\right)}=\dfrac{b}{d}\)
\(\dfrac{4a+3b}{4c+3d}=\dfrac{4bk+3b}{4dk+3d}=\dfrac{b\left(4k+3\right)}{d\left(4k+3\right)}=\dfrac{b}{d}\)
Do đó: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)
3: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3bk+5b}{3bk-5b}=\dfrac{b\left(3k+5\right)}{b\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)
\(\dfrac{3c+5d}{3c-5d}=\dfrac{3dk+5d}{3dk-5d}=\dfrac{d\left(3k+5\right)}{d\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)
Do đó: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
4: \(\dfrac{3a-7b}{b}=\dfrac{3bk-7b}{b}=\dfrac{b\left(3k-7\right)}{b}=3k-7\)
\(\dfrac{3c-7d}{d}=\dfrac{3dk-7d}{d}=\dfrac{d\left(3k-7\right)}{d}=3k-7\)
Do đó: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)
Giải:
Ta có: \(b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c}\)
\(c^2=bd\Rightarrow\frac{b}{c}=\frac{c}{d}\)
\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)
Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\)
\(\Rightarrow a=bk,b=ck,c=dk\)
Ta có:
\(\left(\frac{a+b-c}{b+c-d}\right)^3=\left(\frac{bk+ck-dk}{b+c-d}\right)^3=\left[\frac{k\left(b+c-d\right)}{b+c-d}\right]^3=k^3\) (1)
\(\left(\frac{2a+3b-4c}{2b+3c-4d}\right)^2=\left(\frac{2bk+3ck-4dk}{2b+3c-4d}\right)^3=\left[\frac{k\left(2b+3c-4d\right)}{2b+3c-4d}\right]^3=k^3\) (2)
Từ (1) và (2) suy ra \(\left(\frac{a+b-c}{b+c-d}\right)^3=\left(\frac{2a+3b-4c}{2b+3c-4d}\right)^3\) ( đpcm )
a) Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
Khi đó (2a + 3c)(2b - 3d)
= (2bk + 3dk)(2b - 3d)
= k(2b + 3d)(2b - 3d) (1)
(2a - 3c)(2b + 3d)
= (2bk - 2dk)(2b + 3d)
= k(2b - 3d)(2b + 3d) (2)
Từ (1)(2) => (2a + 3c)(2b - 3d) = (2a - 3c)(2b + 3d)
b) Sửa đề (4a + 3b)(4c - 3d) = (4a - 3b)(4c + 3d)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
Ta có (4a + 3b)(4c - 3d) = (4bk + 3b)(4dk - 3d) = bd(4k + 3)(4k - 3) (1)
Lại có (4a - 3b)(4c + 3d) = (4bk - 3b)(3dk + 3d) = bd(4k- 3)(4k + 3) (2)
Từ (1)(2) => (4a + 3b)(4c - 3d) = (4a - 3b)(4c + 3d)
1, Ta có: \(\frac{a}{b}=\frac{c}{d}\)
\(\Rightarrow\frac{2a}{2b}=\frac{3c}{3d}=\frac{2a+3c}{2b+3d}=\frac{2a-3c}{2b-3d}\)
\(\Rightarrow\left(2a+3c\right).\left(2b-3d\right)=\left(2a-3c\right).\left(2b+3d\right)\)
Vậy (2a + 3c).(2b - 3d) = (2a - 3c).(2b + 3d)
Câu 2 cũng tương tự nên tự làm đi
a, (a-b).(a^2+ab+b^2)
=a(a2+ab+b2)-b(a2+ab+b2)
=a3+a2b+ab2-a2b-ab2-b3
=a3-b3+a2b-a2b+ab2-ab2
=a3-b3
b, (2a- 3b).(4c+d)
=2a.(4c+d)-3b(4c+d)
=8ac+2ad-12bc-3bd