BÀI 1: 1D - 2A - 3C - 4D - 5B - 6C - 7A
BÀI 2: 1B- 2A- 3B - 4B - 5D - 6C - 7A
BÀI 3; 1D - 2C - 3D- 4C - 5B - 6D - 7D - 8D - 9A - 10A - 11D - 12A
BÀI 4: 1D - 2A - 3C - 4A - 5B - 6D - 7A - 8B - 9B - 10A
BÀI 5: 1A - 2D - 3D - 4C - 5B - 6D - 7A
K
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Những câu hỏi liên quan
11 tháng 11 2023
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: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)
DK
2
29 tháng 12 2017
A=-(-2a+3b-5c)+(-5b+4a-2c)-(3c-4b-7a)
=2a-3b+5c-5b+4a-2c-3c+4b+7a
=13a-4b.
18 tháng 6 2019
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
AH
Akai Haruma
Giáo viên
20 tháng 6 2019
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)
\((b+d).c=(b+d).dk=dk(b+d)\)
\((a+c)d=(bk+dk)d=k(b+d)d=dk(b+d)\)
\(\Rightarrow (b+d)c=(a+c)d\)
AH
Akai Haruma
Giáo viên
20 tháng 6 2019
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.
EW
1
29 tháng 12 2017
phá ngoặc lun nà
+4a-5c+3b-2b+a-7c-7b+3c-5a=(4a+a-5a)+(3b-2b-7b)+(-5c-7c+3c)=0-6b-9c=-9c-6b
-2a+3c-b-5b-4c+12a+9b+4c-4a-6a-3b-3c+d=(-2a+12a-4a-6a)+(-b-5b+9b-3b)+(3c-4c+4c-3c)+d=0+0+0+0+d=d
ND
0