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1 a) \(\dfrac{\left(-2\right)}{5}\)= \(\dfrac{-6}{15}\); \(\dfrac{15}{-6}\)= \(\dfrac{5}{-2}\); \(\dfrac{-6}{-2}\)= \(\dfrac{15}{5}\); \(\dfrac{-2}{-6}\)= \(\dfrac{5}{15}\)
Bài 1:
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
a, Ta có: \(\dfrac{a+c}{c}=\dfrac{bk+dk}{dk}=\dfrac{\left(b+d\right)k}{dk}=\dfrac{b+d}{d}\)
\(\Rightarrowđpcm\)
b, Ta có: \(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=\dfrac{k\left(b+d\right)}{b+d}=k\) (1)
\(\dfrac{a-c}{b-d}=\dfrac{bk-dk}{b-d}=\dfrac{k\left(b-d\right)}{b-d}=k\) (2)
Từ (1), (2) \(\Rightarrowđpcm\)
c, Ta có: \(\dfrac{a-c}{a}=\dfrac{bk-dk}{bk}=\dfrac{k\left(b-d\right)}{bk}=\dfrac{b-d}{b}\)
\(\Rightarrowđpcm\)
d, Ta có: \(\dfrac{3a+5b}{2a-7b}=\dfrac{3bk+5b}{2bk-7b}=\dfrac{b\left(3k+5\right)}{b\left(2k-7\right)}=\dfrac{3k+5}{2k-7}\)(1)
\(\dfrac{3c+5d}{2c-7d}=\dfrac{3dk+5d}{2dk-7d}=\dfrac{d\left(3k+5\right)}{d\left(2k-7\right)}=\dfrac{3k+5}{2k-7}\) (2)
Từ (1), (2) \(\Rightarrowđpcm\)
e, Sai đề
f, \(\left(\dfrac{a-b}{c-d}\right)^{2012}=\left(\dfrac{bk-b}{dk-d}\right)^{2012}=\left[\dfrac{b\left(k-1\right)}{d\left(k-1\right)}\right]^{2012}=\dfrac{b^{2012}}{d^{2012}}\)(1)
\(\dfrac{a^{2012}+b^{2012}}{c^{2012}+d^{2012}}=\dfrac{b^{2012}k^{2012}+b^{2012}}{d^{2012}k^{2012}+d^{2012}}=\dfrac{b^{2012}\left(k^{2012}+1\right)}{d^{2012}\left(k^{2012}+1\right)}=\dfrac{b^{2012}}{d^{2012}}\) (2)
Từ (1), (2) \(\Rightarrowđpcm\)
Cái này chỉ cần làm quy tắc nhân chéo là ra rồi nhé :)
a) \(x=\dfrac{-2,6.42}{-12}\)=9,1
b) x = \(\dfrac{2,5.12}{1.5}\) = 20
c) Nhân chéo: 7.(x-1) = 6.(x+5)
<=> 7x - 7 = 6x +30
<=> 7x - 6x = 7 + 30 (chuyển vế)
-> x = 37
d) Nhân chéo: 25x2 = 24.6 = 144
x2 = \(\dfrac{144}{25}\)=5,76
-> x = \(\sqrt{5,76}\) = 2,4
e) Nhân chéo: (x-2)2 = 4.9 = 36
Ta dễ thấy (x-2)2 = 62
-> x-2 = 6 -> x = 6+2 = 8
TICK NHÉ :)
Bài 2 :
Áp dụng theo dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{7}=\dfrac{y}{13}=\dfrac{x+y}{7+13}=\dfrac{40}{20}=2\)
\(\left[{}\begin{matrix}\dfrac{x}{7}=2\Rightarrow x=14\\\dfrac{y}{13}=2\Rightarrow y=36\end{matrix}\right.\)
Vậy .................
Bài 3 :
Bạn cũng áp dụng dãy tỉ số bằng nhau là ra nhé :
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\)
\(\)2) Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{7}=\dfrac{y}{13}=\dfrac{x+y}{7+13}=\dfrac{40}{20}=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=2.7=14\\y=2.13=26\end{matrix}\right.\)
3)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\)
\(\rightarrowđpcm\)
1)
a) \(1\dfrac{5}{6}=\dfrac{-x}{5}\)
\(\Rightarrow\dfrac{11}{6}=\dfrac{-x}{5}\)
\(\Rightarrow-x=\dfrac{5.11}{6}=\dfrac{55}{6}\)
\(\Rightarrow x=-\dfrac{55}{6}\)
b) 4,25 : 8 = -3,5 : x
\(\dfrac{4,25}{8}=\dfrac{-3,5}{x}\)
\(x=\dfrac{-3,5.8}{4,25}\)
\(x=\dfrac{-28}{4,25}\)
2.
\(-\dfrac{12}{1,6}=\dfrac{55}{-7\dfrac{1}{3}}\)
\(\Rightarrow-\dfrac{12}{1,6}=\dfrac{55}{-\dfrac{22}{3}}\)
Ta có thể lặp đc các tỉ lệ thức sau:
\(-\dfrac{12}{1,6}=\dfrac{55}{-\dfrac{22}{3}}\)
\(\dfrac{-\dfrac{22}{3}}{1,6}=\dfrac{55}{-12}\)
\(-\dfrac{12}{55}=\dfrac{1,6}{-\dfrac{22}{3}}\)
\(\dfrac{1,6}{-12}=\dfrac{-\dfrac{22}{3}}{55}\)
1.Tính
a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)
b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)
c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)
d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)
e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)
Bài 2
a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)
\(x=\dfrac{13}{49}\)
b.\(\left|x-1,5\right|=2\)
Xảy ra 2 trường hợp
TH1
\(x-1,5=2\)
\(x=3,5\)
TH2
\(x-1,5=-2\)
\(x=-0,5\)
Vậy \(x=3,5\) hoặc \(x=-0,5\) .
Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.
2.
a) \(\dfrac{x-1}{4}=\dfrac{9}{x-1}\)
\(\Leftrightarrow\left(x-1\right)^2=9.4\)
\(\Leftrightarrow\left(x-1\right)^2=36\)
\(\Leftrightarrow\left(x-1\right)^2=\left[{}\begin{matrix}6^2\\-6^2\end{matrix}\right.\)
\(\Leftrightarrow x-1=\left[{}\begin{matrix}6\\-6\end{matrix}\right.\)
\(\Leftrightarrow x=\left[{}\begin{matrix}7\\-5\end{matrix}\right.\)
1. a. \(\dfrac{3}{-1}=\dfrac{-15}{5}\); \(\dfrac{-1}{3}=\dfrac{5}{-15}\)
\(\dfrac{3}{-15}\)= \(\dfrac{5}{-1}\); \(\dfrac{-15}{3}=\dfrac{-1}{5}\)
b. \(\dfrac{4}{-3}=\dfrac{-12}{9};\dfrac{-3}{4}=\dfrac{9}{-12}\)
\(\dfrac{4}{-12}=\dfrac{9}{-3};\dfrac{-12}{4}=\dfrac{-3}{9}\)
c. \(\dfrac{3}{7}=\dfrac{c}{b};\dfrac{7}{3}=\dfrac{b}{c}\)
\(\dfrac{3}{c}=\dfrac{b}{7};\dfrac{c}{3}=\dfrac{7}{b}\)
d. \(\dfrac{a}{b}=\dfrac{y}{x};\dfrac{b}{a}=\dfrac{x}{y}\)
\(\dfrac{a}{y}=\dfrac{x}{b};\dfrac{y}{a}=\dfrac{b}{x}\)
chúc bạn học tốt
\(xy-3x-y=6\)
\(=>xy+3x-y-3=6-3\)
\(=>x\left(y+3\right)-\left(y+3\right)=3\)
\(=>\left(y+3\right)\left(x-1\right)=3\)
| y+3 | -1 | 3 | 1 | -3 | |
| x-1 | -3 | 1 | 3 | -1 |
| y+3 | -1 | 3 | -3 | 1 |
| y | -4 | -1 | -7 | -3 |
| x-1 | -3 | 1 | 3 | -1 |
| x | -2 | 2 | 4 | 0 |
Mấy bài dễ tự làm nhé:D
1)
Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{bk}{b\left(k+1\right)}=\dfrac{k}{k+1}\\\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{dk}{d\left(k+1\right)}=\dfrac{k}{k+1}\end{matrix}\right.\)
Ta có điều phải chứng minh
\(\left\{{}\begin{matrix}\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\\\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\end{matrix}\right.\)
Ta có điều phải chứng minh