Tìm x, biết:
(x - 6)3 = (x - 6)2
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
bài 1 : a,ta có 3/x-1 =4/y-2=5/z-3 => x-1/3=y-2/4=z-3/5
áp dụng .... => x-1+y-2+z-3 / 3+4+5 = x+y+z-1-2-3/3+4+5 = 12/12=1
do x-1/3 = 1 => x-1 = 3 => x= 4 ( tìm y,z tương tự
Bài 1:
a) Ta có: 3/x - 1 = 4/y - 2 = 5/z - 3 => x - 1/3 = y - 2/4 = z - 3/5 áp dụng ... =>x - 1 + y - 2 + z - 3/3 + 4 + 5 = x + y + z - 1 - 2 - 3/3 + 4 + 5 = 12/12 = 1 do x - 1/3 = 1 => x - 1 = 3 => x = 4 ( tìm y, z tương tự )
\(1,\sqrt{3}x-3=\sqrt{27}\)
\(\Leftrightarrow\sqrt{3}x-3=3\sqrt{3}\)
\(\Leftrightarrow\sqrt{3}\left(x-\sqrt{3}\right)=3\sqrt{3}\)
\(\Leftrightarrow x-\sqrt{3}=3\)
\(\Leftrightarrow x=3+\sqrt{3}\)
\(2,\sqrt{2}x-\sqrt{28}=\sqrt{32}\)
\(\Leftrightarrow\sqrt{2}x-2\sqrt{7}=4\sqrt{2}\)
\(\Leftrightarrow\sqrt{2}x=4\sqrt{2}+2\sqrt{7}\)
\(\Leftrightarrow x=\dfrac{\sqrt{2^2}\left(2\sqrt{2}+\sqrt{7}\right)}{\sqrt{2}}\)
\(\Leftrightarrow x=\sqrt{2}\left(2\sqrt{2}+\sqrt{7}\right)\)
\(\Leftrightarrow x=4+\sqrt{14}\)
\(3,\sqrt{6}x-2\sqrt{6}=\sqrt{54}\)
\(\Leftrightarrow\sqrt{6}\left(x-2\right)=3\sqrt{6}\)
\(\Leftrightarrow x-2=3\)
\(\Leftrightarrow x=5\)
\(4,\sqrt{3}x-\sqrt{2}x=\sqrt{3}+\sqrt{2}\)
\(\Leftrightarrow\left(\sqrt{3}-\sqrt{2}\right)x=\sqrt{3}+\sqrt{2}\)
\(\Leftrightarrow x=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\)
\(\Leftrightarrow x^2-36-x^2+12x-9=9\)
\(\Leftrightarrow12x=54\)
hay x=9/2
Ta có
( x – 6 ) ( x + 6 ) – ( x + 3 ) 2 = 9 ⇔ x 2 – 36 – ( x 2 + 6 x + 9 ) = 9 ⇔ x 2 – 36 – x 2 – 6 x – 9 – 9 = 0
ó - 6x – 54 = 0 ó 6x = -54 ó x = -9
Vậy x = -9
Đáp án cần chọn là: A
3:
a: 3^x*3=243
=>3^x=81
=>x=4
b; 2^x*16^2=1024
=>2^x=4
=>x=2
c: 64*4^x=16^8
=>4^x=4^16/4^3=4^13
=>x=13
d: 2^x=16
=>2^x=2^4
=>x=4
a: Ta có: \(2x\left(x-1\right)-2x^2=-6\)
\(\Leftrightarrow2x^2-2x-2x^2=-6\)
\(\Leftrightarrow x=3\)
b: Ta có: \(2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)
chuyen ve phai sang trai roi dat thua so chung nhe
Bài giải
\(\left(x-6\right)^3=\left(x-6\right)^2\)
\(\Rightarrow\text{ }\left(x-6\right)^3-\left(x-6\right)^2=0\)
\(\Rightarrow\text{ }\left(x-6\right)^2\left[\left(x-6\right)-1\right]=0\)
\(\Rightarrow\text{ }\left(x-6\right)^2\left[x-6-1\right]=0\)
\(\Rightarrow\text{ }\left(x-6\right)^2\left[x-7\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-6\right)^2=0\\x-7=0\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x-6=0\\x=7\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=6\\x=7\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{6\text{ ; }7\right\}\)