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Bài 1:
a) ta có: \(\frac{x-1}{5}=\frac{y-2}{3}=\frac{z-2}{2}=\frac{2y-4}{6}\)
ADTCDTSBN
có: \(\frac{x-1}{5}=\frac{2y-4}{6}=\frac{z-2}{2}=\frac{x-1+2y-4-z+2}{5+6-2}\)\(=\frac{\left(x+2y-z\right)-\left(1+4-2\right)}{9}=\frac{6-3}{9}=\frac{3}{9}=\frac{1}{3}\)
=>...
bn tự tính típ nhé!
b) ta có: \(\frac{x}{y}=\frac{2}{3}\Rightarrow\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x^2}{4}=\frac{y^2}{9}\)
ADTCDTSBN
có: \(\frac{x^2}{4}=\frac{y^2}{9}=\frac{x^2+y^2}{4+9}=\frac{52}{13}=4\)
=>...
Bài 2:
a) ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\)
\(\Rightarrow\frac{b}{d}=\frac{a+b}{c+d}\Rightarrow\frac{a+b}{b}=\frac{c+d}{b}\left(đpcm\right)\)
b) ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{ac}{bd}\) (*)
mà \(\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}\)
Từ (*) \(\Rightarrow\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\left(đpcm\right)\)
Bài 1.
a) Nhân 2 vào tỉ số thứ 2 rồi áp dụng tính chất của dãy tỉ số bằng nhau.
Kết quả:
\(\left\{{}\begin{matrix}x=\dfrac{8}{3}\\y=3\\z=\dfrac{8}{3}\end{matrix}\right.\)
b) \(\dfrac{x}{y}=\dfrac{2}{3}\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{3}\Leftrightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}\)
Theo tính chất dãy tỉ số bằng nhau:
\(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{x^2+y^2}{4+9}=\dfrac{52}{13}=4\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2=16\\y^2=36\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\pm4\\y=\pm6\end{matrix}\right.\)
Vậy ...
Bài 2.
a) \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{b}+1=\dfrac{c}{d}+1\Leftrightarrow\dfrac{a+b}{b}=\dfrac{c+d}{d}\)
b) \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{ac}{bd}=\dfrac{c^2}{d^2}\)
\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{ac}{bd}=\dfrac{a^2}{b^2}\)
\(\Leftrightarrow\dfrac{ac}{bd}=\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}=\dfrac{a^2+c^2}{b^2+d^2}\)
Vậy ...
2:
b) Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=i\Rightarrow\left\{{}\begin{matrix}a=bi\\c=di\end{matrix}\right.\)
Ta có:
\(\dfrac{ac}{bd}=\dfrac{c^2i}{d^2i}=\dfrac{c^2}{d^2}=\left(\dfrac{c}{d}\right)^2=i^2\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2i^2+d^2i^2}{b^2+d^2}=\dfrac{i^2\left(b^2+d^2\right)}{b^2+d^2}=i^2\)
Từ đó suy ra \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\) (đpcm)
4) Ta có: a2=bc => aa=bc =>\(\dfrac{a}{b}=\dfrac{c}{a}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{a}=k\left(k\ne0\right)\)
=> a=bk ; c=ak
+)\(\dfrac{a+b}{a-b}=\dfrac{bk+b}{bk-b}=\dfrac{b\left(k+1\right)}{b\left(k-1\right)}=\dfrac{k+1}{k-1}\left(1\right)\)
+) \(\dfrac{c+a}{c-a}=\dfrac{ak+a}{ak-a}=\dfrac{a\left(k+1\right)}{a\left(k-1\right)}=\dfrac{k+1}{k-1}\left(2\right)\)
Từ (1) và (2) => \(\dfrac{a+b}{a-b}=\dfrac{c+a}{c-a}\)
5) phải xét 2 trường họp dài lắm nên mình chả muốn làm ~~
5a.
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+....+\dfrac{1}{19.21}\\ =\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+....+\dfrac{1}{19}-\dfrac{1}{21}\right)\\ =\dfrac{1}{2}\left(1-\dfrac{1}{21}\right)\\ =\dfrac{1}{2}.\dfrac{20}{21}=\dfrac{10}{21}\)
b.
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\\ =\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+....+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\\ =\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)< \dfrac{1}{2}.1=\dfrac{1}{2}\)
phần a
vì x/2= y/3
y/5= z/4
=>x/2 nhân 1.5 = y/3 nhân 1/5
=> y/5 nhân 1/3 = z/4 nhân 1/3
=>x/10 = y/15 (1)
=>y/15 = z/12 (2)
Từ (1) , (2) ta có :
x/10 = y/15 = z/12
áp dụng t/c......
=>x/10 = y/15 = z/12
=>x+y+z/10+15+12
=> -49/37
b lm tiếp bc tiếp theo nhé✔
Vì mk cmt đầu tiên lên b tích dùm m☢
Câu 1:
Ta có: \(\left[\dfrac{1}{2.5}+\dfrac{1}{5.8}+...+\dfrac{1}{65.68}\right]x-\dfrac{7}{34}=\dfrac{19}{68}\)
\(\Rightarrow\left[\dfrac{1}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{65.68}\right)\right]x=\dfrac{33}{68}\)
\(\Rightarrow\left[\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\right]x=\dfrac{33}{68}\)
\(\Rightarrow\left[\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\right]x=\dfrac{33}{68}\)
\(\Rightarrow\dfrac{11}{68}x=\dfrac{33}{68}\)
\(\Rightarrow x=3\)
Vậy \(x=3.\)
Từ \(a\left(y+z\right)=b\left(z+x\right)\), áp dụng t/c dãy tỉ số bằng nhau ta được
\(\dfrac{z+x}{a}=\dfrac{y+z}{b}=\dfrac{z+x-y-z}{a-b}=\dfrac{x-y}{a-b}\)
\(\Rightarrow\dfrac{z+x}{a}.\dfrac{1}{c}=\dfrac{y+z}{b}.\dfrac{1}{c}=\dfrac{x-y}{c\left(a-b\right)}\)(1)
Tương tự : từ \(b\left(z+x\right)=c\left(x+y\right)\)
\(\Rightarrow\dfrac{z+x}{c}=\dfrac{x+y}{b}=\dfrac{z+x-x-y}{c-b}=\dfrac{y-z}{c-b}\)\(\Rightarrow\dfrac{z+x}{c}.\dfrac{1}{a}=\dfrac{x+y}{b}.\dfrac{1}{a}=\dfrac{y-z}{c-b}.\dfrac{1}{a}\)
\(\Rightarrow\dfrac{z+x}{ac}=\dfrac{x+y}{ab}=\dfrac{y-z}{a\left(c-b\right)}\)(2)
từ \(a\left(y+z\right)=c\left(x+y\right)\)
\(\Rightarrow\dfrac{y+z}{c}=\dfrac{x+y}{a}=\dfrac{y+z-x-y}{c-a}=\dfrac{z-x}{c-a}\)\(\Rightarrow\dfrac{y+z}{c}.\dfrac{1}{b}=\dfrac{x+y}{a}.\dfrac{1}{b}=\dfrac{z-x}{c-a}.\dfrac{1}{b}\)
\(\Rightarrow\dfrac{y+z}{bc}=\dfrac{x+y}{ab}=\dfrac{z-x}{b\left(c-a\right)}\)(3)
Kết hợi (1);(2)(3) => ĐPCM
tik mik nha !!!