![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,5.\left(x+35\right)=515\)
\(x+35=103\)
\(x=103-35\)
\(x=68\)
\(b,12.x-33=3^2.3^3\)
\(12x-33=243\)
\(12x=276\)
\(x=23\)
\(d,12.\left(x-1\right):3=4^3-2^3\)
\(12.\left(x-1\right):3=56\)
\(12.\left(x-1\right)=168\)
\(x-1=14\)
\(x=15\)
![](https://rs.olm.vn/images/avt/0.png?1311)
3.(x-2)+150=240
=>3.(x-2)=90
=>x-2=30
=>x=32
360:(x-7)=90
=>x-7=4
=>x=11
Cau cuối mình không hiểu đề cho lắm
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1: Tìm x,y,z
a) Đặt \(\frac{x}{6}=\frac{y}{5}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=6k\\y=5k\end{matrix}\right.\)
Ta có: \(xy=192\)
\(\Leftrightarrow6k\cdot5k=192\)
\(\Leftrightarrow30k^2=192\)
\(\Leftrightarrow k^2=6.4\)
\(\Leftrightarrow k=\pm\frac{4\sqrt{10}}{5}\)
Ta có: \(\left\{{}\begin{matrix}x=6k\\y=5k\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\cdot\frac{\pm4\sqrt{10}}{5}=\pm\frac{24\sqrt{10}}{5}\\y=5\cdot\pm\frac{4\sqrt{10}}{5}=\pm4\sqrt{10}\end{matrix}\right.\)
Vậy: \(\left(x,y\right)=\left\{\left(\frac{24\sqrt{10}}{5};4\sqrt{10}\right);\left(\frac{-24\sqrt{10}}{5};-4\sqrt{10}\right)\right\}\)
b) Đặt \(\frac{x}{-3}=\frac{y}{7}=a\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-3a\\y=7a\end{matrix}\right.\)
Ta có: \(x^2-y^2=-360\)
\(\Leftrightarrow\left(-3a\right)^2-\left(7a\right)^2=-360\)
\(\Leftrightarrow9a^2-49a^2+360=0\)
\(\Leftrightarrow360-40a^2=0\)
\(\Leftrightarrow40a^2=360\)
\(\Leftrightarrow a^2=9\)
hay \(a=\pm3\)
Trường hợp 1: a=3
\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\cdot3\\y=7\cdot3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-9\\y=21\end{matrix}\right.\)
Trường hợp 2: a=-3
\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\cdot\left(-3\right)\\y=7\cdot\left(-3\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=9\\y=-21\end{matrix}\right.\)
Vậy: (x,y)={(-9;21);(9;-21)}
c) Ta có: \(\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{4}\)
\(\Leftrightarrow\frac{x-1}{2}=\frac{2y+4}{6}=\frac{3z-9}{12}\)
mà x-2y+3z=46
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\frac{x-1}{2}=\frac{2y+4}{6}=\frac{3z-9}{12}=\frac{x-1-2y-4+3z-9}{2-6+12}=\frac{46-14}{8}=\frac{32}{8}=4\)
Do đó:
\(\left\{{}\begin{matrix}x-1=4\cdot2=8\\2y+4=4\cdot6=24\\3z-9=4\cdot12=48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=9\\2y=20\\3z=57\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=9\\y=10\\z=19\end{matrix}\right.\)
Vậy: (x,y,z)=(9;10;19)
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ 34 . 3n : 9 = 34 => 34 . 3n = 34 x 9 => 34 . 3n = 306 => 3n = 306 : 34 => 3n = 9 => n = 2
b/ 9 < 3n < 27 => 32 < 3n < 33 => 2 < n < 3
Mà: n thuộc N => n không tồn tại
c/ Chữ số tận cùng của 360 là 0
d/ Ta có: A = 1 + 3 + 32 + 33 + 34 + 35 + 36
=> 3A = 3 + 32 + 33 + 34 + 35 + 36 + 37
=> 3A - A = 2A = (3 + 32 + 33 + 34 + 35 + 36 + 37) - (1 + 3 + 32 + 33 + 34 + 35 + 36 ) = 3 + 32 + 33 + 34 + 35 + 36 + 37 - 1 - 3 - 32 - 33 - 34 - 35 - 36
=> 2A = 37 - 1 => A = (37 - 1) : 2 < 37 - 1 = B
=> A < B
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
\(\left(-\dfrac{72}{40}-\dfrac{144}{60}-2\dfrac{1}{3}\right):\left(\dfrac{45}{100}-\dfrac{25}{60}+-\dfrac{75}{25}\right)\)
\(=\left(-\dfrac{9}{5}-\dfrac{12}{5}-\dfrac{7}{3}\right):\left(\dfrac{9}{20}-\dfrac{5}{12}+-3\right)\)
\(=\left(-\dfrac{27}{15}-\dfrac{36}{15}-\dfrac{21}{15}\right):\left(\dfrac{27}{60}-\dfrac{25}{60}+-3\right)\)
\(=\left(-\dfrac{28}{5}\right):\left(-\dfrac{89}{30}\right)\)
\(=\left(-\dfrac{28}{5}\right).\left(-\dfrac{30}{89}\right)\)
\(=\dfrac{168}{89}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(-5\right)\left(x-2\right)^2+360=\left(-150\right)\cdot3+43\cdot5\)
\(-5\cdot\left(x-2\right)^2+360=-235\)
\(-5\cdot\left(x-2\right)^2=-595\)
\(\left(x-2\right)^2=119\)
\(\left(x-2\right)^2=\left(\pm\sqrt{199}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}x-2=\sqrt{119}\\x-2=-\sqrt{119}\end{cases}\Rightarrow\orbr{\begin{cases}x=\sqrt{119}+2\\x=2-\sqrt{119}\end{cases}}}\)
b) \(\left(x+5\right)-\left(3x+9\right)=-16\)
\(x+5-3x-9=-16\)
\(-2x-4=-16\)
\(-2x=-12\)
\(x=6\)
c) \(3\left(x+2\right)-\left(15-x\right)\cdot6=160+\left(-1\right)^{1005}\)
\(3x+6-90+6x=160-1\)
\(9x-84=159\)
\(9x=243\)
\(x=27\)
d) \(x\left(x-1\right)\left(x^2-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x\left(x-1\right)=0\\x^2-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\left\{0;1\right\}\\x=\left\{2;-2\right\}\end{cases}}}\)
= 3
bài gì ?
3^3+x + 3^2+x + 3^1+x =360
3x +3^3 + 3^2 + 3^1 =360
3x +27+9+3 =360
3x +39 =360
3x =360-39
3x =321
3x = 107