câu 8
a) 3,45*x +6,55*x=20%
b)50%/x=3+2 và 2/3
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a: =>10x=0,2
=>x=0,02
b: =>1/2:x=3-2,5=0,5
=>0,5:x=0,5
=>x=1
3,45 \(\times\) \(x\) + 6,55 \(\times\)\(x\) = 20%
\(x\) \(\times\) ( 3,45 + 6,55) = 0,2
\(x\) \(\times\) 10 = 0,2
\(x\) = 0,2 : 10
\(x\) = 0,02
a) 1/3 * x + 2/5 = 3/2
1/3 * x = 3/2 - 2/5
1/3 * x = 15/10 - 4/10
1/3 * x = 11/10
x = 11/10 : 1/3
x = 11/10 * 3
x = 33/10
\(\left(3,45+6,55\right)\times x=10\)
\(\Leftrightarrow10x=10\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)
\(3,45\times x+6,55\times x=2,5\)
\(\left(3,45+6,55\right)\times x=2,5\)
\(10\times x=2,5\)
x = 0 , 25
\(3,45\times x+6,55\times x=2,5\)
\(x\times\left(3,45+6,55\right)=2,5\)
\(x\times10=2,5\)
\(x=2,5:10\)
\(x=0,25\)
3,45 x a + 6,55 x a = 2,5
a x ( 3,45 + 6,55 ) = 2,5
a x 10 = 2,5
a = 2,5 : 10
a = 0,25
Đổi 2,5 m = 25 dm 1,5 m = 15 dm
Chiều cao của bể là :
6750 : 25 : 15 = 18 ( dm )
Đáp số : 18 dm
Câu 1:
= 3,45 x ( 123 + 4 - 27 )
= 345
Câu 2:
3 x x = 3,45 : 5
3 x x = 0,69
x= 0,69 : 3 = 0,23
Nhớ k cho mình
câu 2 (3*x)*5 = 3.45
3xx =3,45/5
3xx =0,69
3xx =0,69/3
x =0,23
a) \(\sqrt{\left(2x-3\right)^2}=7\)
\(\Leftrightarrow\left|2x-3\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
b) \(\sqrt{64x+128}-\sqrt{25x+50}+\sqrt{4x+8}=20\left(đk:x\ge-2\right)\)
\(\Leftrightarrow8\sqrt{x+2}-5\sqrt{x+2}+2\sqrt{x+2}=20\)
\(\Leftrightarrow5\sqrt{x+2}=20\)
\(\Leftrightarrow\sqrt{x+2}=4\Leftrightarrow x+2=16\Leftrightarrow x=14\left(tm\right)\)
c) \(\sqrt{x^2-9}-3\sqrt{x-3}=0\left(đk:x\ge3\right)\)
\(\Leftrightarrow\sqrt{\left(x-3\right)\left(x+3\right)}-3\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\\sqrt{x+3}=3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x+3=9\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)
a. \(\sqrt{\left(2x-3\right)^2}=7\)
<=> \(\left|2x-3\right|=7\)
<=> \(\left[{}\begin{matrix}2x-3=7\left(x\ge\dfrac{3}{2}\right)\\-2x+3=7\left(x< \dfrac{3}{2}\right)\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}2x=10\\-2x=4\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=5\left(TM\right)\\x=-2\left(TM\right)\end{matrix}\right.\)
b. \(\sqrt{64x+128}-\sqrt{25x+50}+\sqrt{4x+8}=20\) ĐK: \(x\ge-2\)
<=> \(\sqrt{64\left(x+2\right)}-\sqrt{25\left(x+2\right)}+\sqrt{4\left(x+2\right)}-20=0\)
<=> \(8\sqrt{x+2}-5\sqrt{x+2}+2\sqrt{x+2}-20=0\)
<=> \(\sqrt{x+2}.\left(8-5+2\right)-20=0\)
<=> \(5\sqrt{x+2}=20\)
<=> \(\sqrt{x+2}=4\)
<=> \(\left(\sqrt{x+2}\right)^2=4^2\)
<=> \(\left|x+2\right|=16\)
<=> \(\left[{}\begin{matrix}x+2=16\left(x\ge-2\right)\\x+2=-16\left(x< -2\right)\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=14\left(TM\right)\\x=-18\left(TM\right)\end{matrix}\right.\)
c. \(\sqrt{x^2-9}-3\sqrt{x-3}=0\) ĐK: \(x\ge3\)
<=> \(\sqrt{\left(x-3\right)\left(x+3\right)}-3\sqrt{x-3}=0\)
<=> \(\sqrt{x-3}.\sqrt{x+3}-3\sqrt{x-3}=0\)
<=> \(\left(\sqrt{x+3}-3\right).\sqrt{x-3}=0\)
<=> \(\left[{}\begin{matrix}\sqrt{x+3}-3=0\\\sqrt{x-3}=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=6\\x=3\end{matrix}\right.\)
a, 3,45 \(\times\) \(x\) + 6,55 \(\times\) \(x\) = 20%
\(x\) \(\times\) ( 3,45 + 6,55) = 0,2
\(x\) \(\times\) 10 = 0,2
\(x\) = 0,2 : 10
\(x\) = 0,02
b, \(\dfrac{50\%}{x}\) = 3 + 2\(\dfrac{2}{3}\)
0,5: \(x\) = 3 + \(\dfrac{8}{3}\)
0,5: \(x\) = \(\dfrac{17}{3}\)
\(x\) = 0,5 : \(\dfrac{17}{3}\)
\(x\) = \(\dfrac{3}{34}\)