Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Vì \(b\ne d;b+d\ne0\) nên áp dụng tính chất cảu dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\)
Vậy \(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\) (đpcm)
Chúc bạn học tốt!!!
Ta có:Nếu
\(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\)
thì \((a+c)(b-d)=(a-c)(b+d)\)
\(a(b-d)+c(b-d)=a(b+d)-c(b+d)\)
\(ab-ad+bc-cd=ab+ad-bc+cd\)
\(=\)\(ab-ab\)\(-ad+ad\)\(+bc-bc\)\(-cd+cd\)
\(=0\)
\(\Leftrightarrow\left(a+c\right)\left(b-d\right)\)\(=\left(a-c\right)\left(b+d\right)\)
\(\Leftrightarrow\dfrac{a+c}{b+d}\)\(=\dfrac{a-c}{b-d}\)
a)\(\dfrac{-8}{-2}=\dfrac{-4}{-1}=\dfrac{4}{1}=\dfrac{8}{2}=\dfrac{12}{3}=4\)
Vậy hai đại lượng x và y ở bảng a) tỉ lệ thuận với nhau.
b) \(\dfrac{22}{1}\ne\dfrac{100}{5}\)
Vậy hai đại lượng x và y ở bảng b không tỉ lệ thuận với nhau.
\(a,x^2-113=31\\ \Leftrightarrow x^2=144\\ \Leftrightarrow x=\pm12\\ Vay...\\ b,\sqrt{x+2,29}=2.3\\ \Leftrightarrow x+2,29=6^2\\ x=36-2,29=33,71\\ c,x^4=256\\ \Leftrightarrow x=\pm4\\ Vay...\\ d,\left(\sqrt{x}-1\right)^2=0,5625\\ \Leftrightarrow\sqrt{x}-1\in\left\{-0,75;0,75\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{0,25;1,75\right\}\\ Vay...\\ e,2\sqrt{x}-x=0\\ \Leftrightarrow\sqrt{x}\left(2-\sqrt{x}\right)=0\\ \Leftrightarrow\sqrt{x}=0hoac2-\sqrt{x}=0\\ \Leftrightarrow x=0hoacx=4\\ f,x+\sqrt{x}=0\\ \Leftrightarrow\sqrt{x}\left(\sqrt{x}+1\right)=0\\ \Leftrightarrow x=0hoacx=1\)
a. x2−113=31
=> x2=144
=> x2=\(\sqrt{144}\)
=> x=\(\pm12\)
c.x4=256
=> x4=44
=> x=\(\pm4\)
>> Mình không chép lại đề bài nhé ! <<
Cách 1 :
\(A=\left(\dfrac{36-4+3}{6}\right)-\left(\dfrac{30+10-9}{6}\right)-\left(\dfrac{18-14+15}{6}\right)=\dfrac{35}{6}-\dfrac{31}{6}-\dfrac{19}{6}=-\dfrac{15}{6}=-\dfrac{5}{2}\)
Cách 2 :
\(A=6-\dfrac{2}{3}+\dfrac{1}{2}-5+\dfrac{5}{3}-\dfrac{3}{2}-3-\dfrac{7}{3}+\dfrac{5}{2}\)
\(A=\left(6-5-3\right)-\left(\dfrac{2}{3}+\dfrac{5}{3}-\dfrac{7}{3}\right)+\left(\dfrac{1}{2}+\dfrac{3}{2}-\dfrac{5}{2}\right)\)
\(A=-2-0-\dfrac{1}{2}=-\dfrac{5}{2}\)
Cách 1 :
\(\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)
\(=\left(\dfrac{36}{6}-\dfrac{4}{6}+\dfrac{3}{6}\right)-\left(\dfrac{30}{6}+\dfrac{10}{6}-\dfrac{9}{6}\right)-\left(\dfrac{18}{6}-\dfrac{14}{6}+\dfrac{15}{6}\right)\)
\(=\dfrac{35}{6}-\dfrac{31}{6}-\dfrac{19}{6}\)
\(=-\dfrac{5}{2}\)
Cách 2 :
\(\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)
\(=6-\dfrac{2}{3}+\dfrac{1}{2}-5-\dfrac{5}{3}+\dfrac{3}{2}-3+\dfrac{7}{3}-\dfrac{5}{2}\)
\(=\left(6-5-3\right)+\left(\dfrac{-2}{3}+\dfrac{-5}{3}+\dfrac{7}{3}\right)+\left(\dfrac{1}{2}+\dfrac{3}{2}+\dfrac{-5}{2}\right)\)
\(=\left(-2\right)+0+\dfrac{-1}{2}\)
\(=\dfrac{-5}{2}\)
Bài 1:
a)\(\frac{2}{3}.\frac{5}{2}-\frac{3}{4}.\frac{2}{3}=\frac{5}{3}-\frac{1}{2}=\frac{7}{6}\)
b)\(2.\left(\frac{-3}{2}\right)^2-\frac{7}{2}=\frac{2.9}{4}-\frac{7}{2}=\frac{9-7}{2}=\frac{2}{2}=1\)
c)\(-\frac{3}{4}.\frac{68}{13}-0,75.\frac{36}{13}=\frac{-3.4.17}{4.13}-\frac{3.9.4}{4.13}=\frac{-51-27}{13}=\frac{-78}{13}=-6\)
Bài 2:
a)|x-1,4|=1,6
\(\Rightarrow\left[\begin{array}{nghiempt}x-1,4=1,6\\x-1,4=-1,6\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=3\\x=-0,2\end{array}\right.\)
b) \(\frac{3}{4}-x=\frac{4}{5}\)
\(x=\frac{3}{4}-\frac{4}{5}=-\frac{1}{20}\)
c)(1-2x)3=-8
(1-2x)3=(-2)3
1-2x=-2
2x=3
x=\(\frac{3}{2}\)
Bài 3:
\(\frac{x}{2}=\frac{y}{5}=\frac{z}{7}=k\)
\(\Rightarrow\begin{cases}x=2k\\y=5k\\z=7k\end{cases}\)
A=\(\frac{2k-5k+7k}{2k+2.5k-7k}=\frac{4k}{5k}=\frac{4}{5}\)
=> x=4/5 . 2= 8/5
y=4/5 . 5=4
z=4/5.7=28/5
lỗi part 2
y=-10