Tìm x biết \(3.|x|-x=6\)
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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 )
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\(\Leftrightarrow x^2-36-x^2+12x-9=9\)
\(\Leftrightarrow12x=54\)
hay x=9/2
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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
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\(1,\sqrt{3}x-3=\sqrt{27}\)
\(\Leftrightarrow\sqrt{3}x-3=3\sqrt{3}\)
\(\Leftrightarrow\sqrt{3}\left(x-\sqrt{3}\right)=3\sqrt{3}\)
\(\Leftrightarrow x-\sqrt{3}=3\)
\(\Leftrightarrow x=3+\sqrt{3}\)
\(2,\sqrt{2}x-\sqrt{28}=\sqrt{32}\)
\(\Leftrightarrow\sqrt{2}x-2\sqrt{7}=4\sqrt{2}\)
\(\Leftrightarrow\sqrt{2}x=4\sqrt{2}+2\sqrt{7}\)
\(\Leftrightarrow x=\dfrac{\sqrt{2^2}\left(2\sqrt{2}+\sqrt{7}\right)}{\sqrt{2}}\)
\(\Leftrightarrow x=\sqrt{2}\left(2\sqrt{2}+\sqrt{7}\right)\)
\(\Leftrightarrow x=4+\sqrt{14}\)
\(3,\sqrt{6}x-2\sqrt{6}=\sqrt{54}\)
\(\Leftrightarrow\sqrt{6}\left(x-2\right)=3\sqrt{6}\)
\(\Leftrightarrow x-2=3\)
\(\Leftrightarrow x=5\)
\(4,\sqrt{3}x-\sqrt{2}x=\sqrt{3}+\sqrt{2}\)
\(\Leftrightarrow\left(\sqrt{3}-\sqrt{2}\right)x=\sqrt{3}+\sqrt{2}\)
\(\Leftrightarrow x=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\)
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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
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\(\text{(x+2)(y-3)=5 }\)
\(\Rightarrow\)x+2;y-3\(\in\)Ư(5)
Mà Ư(5)={1;5;-1;-5}
Có bảng:
Th1:
x+2=1;y-3=6
=>x=-3
y=9
Tương tự 3 trường hợp còn lại
|x|-x=6/3=2
x=-1