Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left\{{}\begin{matrix}a=\dfrac{35}{49}=\dfrac{5}{7}\\b=\sqrt{\dfrac{5^2}{7^2}}=\dfrac{5}{7}\\c=\dfrac{\sqrt{5^2}+\sqrt{35^2}}{\sqrt{7^2}+\sqrt{49^2}}=\dfrac{5+35}{7+49}=\dfrac{5}{7}\\d=\dfrac{\sqrt{5^2}-\sqrt{35^2}}{\sqrt{7^2}-\sqrt{49^2}}=\dfrac{5-35}{7-49}=\dfrac{5}{7}\end{matrix}\right.\)
\(\Rightarrow a=b=c=d=\dfrac{5}{7}\)
\(a=\dfrac{35}{49};b=\dfrac{5}{7}\\ c,=\dfrac{5+35}{7+49}=\dfrac{12}{14}=\dfrac{6}{7}\\ d,=\dfrac{5-35}{7-49}\)
Áp dụng t/c dtsbn:
\(\dfrac{5}{7}=\dfrac{35}{49}=\dfrac{5+35}{7+49}=\dfrac{5-35}{7-49}\) hay \(a=b=c=d\)
a) \(\dfrac{49}{81}=\dfrac{7^x}{9^x}\)(sửa đề)
\(\Leftrightarrow\left(\dfrac{7}{9}\right)^2=\left(\dfrac{7}{9}\right)^x\)\(\Rightarrow x=2\)
b) \(\dfrac{-64}{343}=\left(-\dfrac{4^x}{7^x}\right)\)(sửa đề)
\(\Leftrightarrow\left(-\dfrac{4}{7}\right)^3=\left(-\dfrac{4}{7}\right)^x\) \(\Rightarrow x=3\)
c) \(\dfrac{9}{144}=\dfrac{3^x}{12^x}\)(sửa đề)
\(\Leftrightarrow\left(\dfrac{3}{12}\right)^2=\left(\dfrac{3}{12}\right)^x\Rightarrow x=2\)
d) \(-\dfrac{1}{32}=\left(-\dfrac{1^x}{2^x}\right)\)(sửa đề)
\(\Leftrightarrow\left(-\dfrac{1}{2}\right)^5=\left(-\dfrac{1}{2}\right)^x\Rightarrow x=5\)
Mong bạn xem lại đề bài.
a) (x-2).(x+1)=6
(x+1-3).(x+1)=6
(x+1).(1-3)=6
(x+1).-2=6
x+1=6:(-2)
x+1=-3
x=-3-1
x=-4
a) Xét ước
b)
\(\left(x^2+7\right)\left(x^2-49\right)< 0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2+7< 0\Rightarrow x^2< -7\left(KTM\right)\\x^2-49>0\Rightarrow x^2>49\end{matrix}\right.\\\left\{{}\begin{matrix}x^2+7>0\Rightarrow x^2>-7\\x^2-49< 0\Rightarrow x^2< 9=49\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow-7< x^2< 49\)
\(\Rightarrow0\le x^2< 49\)
\(\Rightarrow x^2\in\left\{0;1;4;9;16;25;36\right\}\)
\(\Rightarrow x\in\left\{0;\pm1;\pm2;\pm3;\pm4;\pm5;\pm6\right\}\)
\((x^2-7)(x^2-49)<0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2-7< 0\Rightarrow x^2< 7\\x^2-49>0\Rightarrow x^2>49\end{matrix}\right.\\\left\{{}\begin{matrix}x^2-7>0\Rightarrow x^2>7\\x^2-49< 0\Rightarrow x^2< 49\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow7< x^2< 49\)
\(\Rightarrow x^2\in\left\{9;16;25;36\right\}\)
\(\Rightarrow x\in\left\{\pm3;\pm4;\pm5;\pm6\right\}\)
a) ADTCDTSBN
có: \(\frac{x}{2}=\frac{z}{4}=\frac{x+z}{2+4}=\frac{18}{6}=3.\)
=> x/2 = 3 => x = 6
y/3 = 3 => y = 9
z/4 = 3 => z = 12
KL:...
b,c làm tương tự nha
d) ta có: \(\frac{x}{5}=\frac{y}{-6}=\frac{z}{7}=\frac{2x}{10}\)
ADTCDTSBN
có: \(\frac{2x}{10}=\frac{y}{-6}=\frac{z}{7}=\frac{2x+y-z}{10+\left(-6\right)-7}=\frac{49}{-3}\)
=>...
e) ADTCDTSBN
có: \(\frac{x+1}{2}=\frac{y+2}{3}=\frac{z+3}{4}=\frac{x+1+y+2+z+3}{2+3+4}=\frac{\left(x+y+z\right)+\left(1+2+3\right)}{9}\)
\(=\frac{21+6}{9}=\frac{27}{9}=3\)
=>...
g) ta có: \(\frac{x}{4}=\frac{y}{3}=k\Rightarrow\hept{\begin{cases}x=4k\\y=3k\end{cases}}\)
mà xy = 12 => 4k.3k = 12
12.k2 = 12
k2 = 1
=> k = 1 hoặc k = -1
=> x = 4.1 = 4
y = 3.1 = 3
x=4.(-1) = -4
y=3.(-1) = -3
KL:...
h) ta có: \(\frac{x}{5}=\frac{y}{3}\Rightarrow\frac{x^2}{25}=\frac{y^2}{9}\)
ADTCDTSBN
có: \(\frac{x^2}{25}=\frac{y^2}{9}=\frac{x^2-y^2}{25-9}=\frac{16}{16}=1\)
=>...
\(\frac{x}{7}+\frac{9}{49}=\frac{3}{14}\)
\(\Rightarrow\frac{x}{7}=\frac{3}{14}-\frac{9}{49}=\frac{3}{98}\)
\(\Rightarrow x=3.7:98=\frac{3}{14}\)
Vậy \(x=\frac{3}{14}\)