Cho \(\dfrac{a}{b}=\dfrac{c}{d}\). Chứng minh:
1) \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)
2) \(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)
3) \(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
4) \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)
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18 tháng 6 2019 lúc 16:07
Chứng minh : dfrac{a}{b}dfrac{c}{d} nếu biết :
a,dfrac{4a-3b}{4c-3d}dfrac{4a+3b}{4c+3d}
b,dfrac{2a-3b}{2a+3b}dfrac{2c-3d}{2c+3d}
c,dfrac{3a+5b}{3a-5b}dfrac{3c+5d}{3c-5d}
d,dfrac{4a-3b}{a}dfrac{4c-3d}{c}
e,dfrac{3a-7b}{b}dfrac{3c-7d}{d}
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Chứng minh : \(\dfrac{a}{b}=\dfrac{c}{d}\) nếu biết :
a,\(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)
b,\(\dfrac{2a-3b}{2a+3b}=\dfrac{2c-3d}{2c+3d}\)
c,\(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
d,\(\dfrac{4a-3b}{a}=\dfrac{4c-3d}{c}\)
e,\(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)
a) Có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{4a}{3b}=\frac{4c}{3d}\)
Áp dụng tỉ lệ thức ta có :
\(\frac{4a}{3b}=\frac{4c}{3d}\Rightarrow\)\(\frac{4a}{4c}=\frac{3b}{3d}\Rightarrow\frac{4a+3b}{4c+3d}=\frac{4c-3d}{4c-3d}\)
b) Có : \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{2a}{3b}=\frac{2c}{3d}\)
Áp dụng tỉ lệ thức ta có "
\(\frac{2a}{3b}=\frac{2c}{3d}\Rightarrow\frac{2a}{2c}=\frac{3b}{3d}\Rightarrow\frac{2a-3b}{2c-3d}=\frac{2a3b}{2c+3d}\Rightarrow\frac{2a-3b}{2a+3b}=\frac{2c-3d}{2c+3d}\)
Các câu còn lại bạn làm tương tự
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18 tháng 6 2019 lúc 16:19
Chứng minh dfrac{a}{b}dfrac{c}{d} nếu biết :
a,dfrac{4a-3b}{4c-3d}dfrac{4a+3b}{4c+3d}
b,dfrac{2a-3b}{2a+3b}dfrac{2c-3d}{2c+3d}
c,dfrac{3a+5b}{3a-5b}dfrac{3c+5d}{3c-5d}
d,dfrac{4a-3b}{a}dfrac{4c-3d}{c}
e,dfrac{3a-7b}{b}dfrac{3c-7d}{d}
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Chứng minh \(\dfrac{a}{b}=\dfrac{c}{d}\) nếu biết :
a,\(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)
b,\(\dfrac{2a-3b}{2a+3b}=\dfrac{2c-3d}{2c+3d}\)
c,\(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
d,\(\dfrac{4a-3b}{a}=\dfrac{4c-3d}{c}\)
e,\(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)
1 tháng 11 2021 lúc 14:09
cho \(\dfrac{a}{b}\) = \(\dfrac{c}{d}\). Chứng minh \(\dfrac{2a+3c}{3a+4c}=\dfrac{2b+3d}{3b+4d}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\dfrac{2a+3c}{3a+4c}=\dfrac{2bk+3dk}{3bk+4dk}=\dfrac{2b+3d}{3b+4d}\)
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Cho tỉ lệ thức \(\dfrac{a}{b}\) = \(\dfrac{c}{d}\) . Chứng minh đẳng thức sau : \(\dfrac{2a+3b}{3a-5b}\) = \(\dfrac{2c+3d}{3c-5d}\)
Lời giải:
Đặt $\frac{a}{b}=\frac{c}{d}=k$
$\Rightarrow a=bk, c=dk$
Khi đó:
$\frac{2a+3b}{3a-5b}=\frac{2bk+3b}{3bk-5b}=\frac{b(2k+3)}{b(3k-5)}=\frac{2k+3}{3k-5}(1)$
$\frac{2c+3d}{3c-5d}=\frac{2dk+3d}{3dk-5d}=\frac{d(2k+3)}{d(3k-5)}=\frac{2k+3}{3k-5}(2)$
Từ $(1); (2)$ ta có đpcm.
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Cho a+b+c+d ≠ 0 thỏa mãn:
\(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{b+a+d}=\dfrac{d}{c+b+a}\)
Tính P = \(\dfrac{2a+5b}{3c+4d}+\dfrac{2b+5c}{3d+4a}+\dfrac{2c+5d}{3a+4b}+\dfrac{2d+5a}{3c+4b}\)
Cho a+b+c+d ≠ 0 và \(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{b+a+d}=\dfrac{d}{c+b+a}\)
Tính giá trị biểu thức:
P = \(\dfrac{2a+5b}{3c+4d}-\dfrac{2b+5c}{3d+4a}+\dfrac{2c+5d}{3a+4b}+\dfrac{2d+5a}{3c+4b}\)
16 tháng 6 2019 lúc 9:26
Cho dfrac{a}{b} dfrac{c}{d} . Chứng minh :
a, (a+c).((b-d)(a-c).(b-d)
b, (a+c).b(b+d).a
c, a.(b-d)b(a-c)
d, (b+d).c(a+c).d
e, (b-d).c(a-c).d
f, (a+b).(c-d)(a-b).(c+d)
g, (2a+3c).(2b-3d)(2a-3c).(2b+3d)
h, (4a+3b).(4c-3d)(4a-3b).((4c+3d)
i, (2a+3b).(4c-5d)(4a-5b).(2c+3d)
k, (4a+5b).(7c-11d)(7a-11b).(4c+5d)
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Cho \(\dfrac{a}{b} = \dfrac{c}{d}\) . Chứng minh :
a, \((a+c).((b-d)=(a-c).(b-d)\)
b, \((a+c).b=(b+d).a\)
c, \(a.(b-d)=b(a-c)\)
d, \((b+d).c=(a+c).d\)
e, \((b-d).c=(a-c).d\)
f, \((a+b).(c-d)=(a-b).(c+d)\)
g, \((2a+3c).(2b-3d)=(2a-3c).(2b+3d)\)
h, \((4a+3b).(4c-3d)=(4a-3b).((4c+3d)\)
i, \((2a+3b).(4c-5d)=(4a-5b).(2c+3d)\)
k, \((4a+5b).(7c-11d)=(7a-11b).(4c+5d)\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk,c=dk\). Khi đó ta có:
a)
\((a+c)(b-d)=(bk+dk)(b-d)=k(b+d)(b-d)\)
\((a-c)(b+d)=(bk-dk)(b+d)=k(b-d)(b+d)=k(b+d)(b-d)\)
\(\Rightarrow (a+c)(b-d)=(a-c)(b+d)\) (đpcm)
b)
\((a+c)b=(bk+dk)b=k(b+d).b=bk(b+d)\)
\((b+d).a=(b+d).bk=bk(b+d)\)
\(\Rightarrow (a+c)b=(b+d)a\)
c)
\(a(b-d)=bk(b-d)\)
\(b(a-c)=b(bk-dk)=bk(b-d)\)
\(\Rightarrow a(b-d)=b(a-c)\)
d)
\((b+d).c=(b+d).dk=dk(b+d)\)
\((a+c)d=(bk+dk)d=k(b+d)d=dk(b+d)\)
\(\Rightarrow (b+d)c=(a+c)d\)
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e)
\((b-d).c=(b-d).dk=dk(b-d)\)
\((a-c)d=(bk-dk)d=k(b-d)d=dk(b-d)\)
\(\Rightarrow (b-d)c=(a-c)d\)
f)
\((a+b)(c-d)=(bk+b)(dk-d)=b(k+1)d(k-1)=bd(k-1)(k+1)\)
\((a-b)(c+d)=(bk-b)(dk+d)=b(k-1)d(k+1)=bd(k-1)(k+1)\)
\(\Rightarrow (a+b)(c-d)=(a-b)(c+d)\)
g)
\((2a+3c)(2b-3d)=(2bk+3dk)(2b-3d)=k(2b+3d)(2b-3d)\)
\((2a-3c)(2b+3d)=(2bk-3dk)(2b+3d)=k(2b-3d)(2b+3d)\)
\(\Rightarrow (2a+3c)(2b-3d)=(2a-3c)(2b+3d)\)
h)
\((4a+3b)(4c-3d)=(4bk+3b)(4dk-3d)=b(4k+3)d(4k-3)=bd(4k+3)(4k-3)\)
\((4a-3b)(4c+3d)=(4bk-3b)(4dk+3d)=b(4k-3)d(4k+3)=bd(4k+3)(4k-3)\)
\(\Rightarrow (4a+3b)(4c-3d)=(4a-3b)(4c+3d)\)
i,k: Hoàn toàn tương tự.
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Cho a+b+c+d khác 0 sao cho: \(\dfrac{b+c+d}{a}=\dfrac{a+c+d}{b}=\dfrac{b+a+d}{c}=\dfrac{c+b+a}{d}\)
Hãy tính: M = \(\dfrac{2a+5b}{3c+4d}-\dfrac{2b+5c}{3d+4a}-\dfrac{2c+5d}{3a+4b}+\dfrac{2d+5a}{3c+4b}\)
8 tháng 8 2023 lúc 21:30
Cho \(\dfrac{2a+3c}{2b+3d}\)=\(\dfrac{2a-3c}{2b-3d}\). Chứng minh\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)
\(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}=\dfrac{2a+3c+2a-3c}{2b+3d+2b-3d}=\dfrac{a}{b}\)
\(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}=\dfrac{2a+3c-\left(2a-3c\right)}{2b+3d-\left(2b-3d\right)}=\dfrac{c}{d}\)
Suy ra \(\dfrac{a}{b}=\dfrac{c}{d}\)
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