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`#040911`
`a)`
`6 \times x - 5 = 613`
`=> 6 \times x = 613 + 5`
`=> 6 \times x = 618`
`=> x = 618 \div 6`
`=> x = 103`
Vậy, `x = 103`
`b)`
`12 \times x + 3 \times x = 30`
`=> x \times (12 + 3) = 30`
`=> x \times 15 = 30`
`=> x = 30 \div 15`
`=> x = 2`
Vậy, `x = 2`
`c)`
`125 - 25 \times (x - 1) = 100`
`=> 25 \times (x - 1) = 125 - 100`
`=> 25 \times (x - 1) = 25`
`=> x - 1 = 25 \div 25`
`=> x - 1 = 1`
`=> x = 1 + 1`
`=> x = 2`
Vậy, `x = 2`
`d)`
`(x - 2) \times (9x - 4) = 0?`
`=>`
TH1: `x - 2 = 0`
`=> x = 0 + 2`
`=> x = 2`
TH2: `9x - 4 = 0`
`=> 9x = 4`
`=> x = 4/9`
Vậy, `x \in {2; 4/9}.`
\(a,6x-5=613\\ \Leftrightarrow6x=618\\ \Leftrightarrow x=103\\ b,12x+3x=30\\ \Leftrightarrow15x=30\\ \Leftrightarrow x=2\\ c,125-25\left(x-1\right)=100\\ \Leftrightarrow25\left(x-1\right)=25\\ \Leftrightarrow x-1=1\\ \Leftrightarrow x=2\\ d,\left(x-2\right)\cdot\left(9x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{4}{9}\end{matrix}\right.\)
1.a) 4.(12+88) = 4.100=400
b)72- 6 .[2.8:8 ]=72-6.[16:8]=72-6 .2= 49-12=37 ( tui ko chắc lắm )
2.a) x=26-11=15
b) (4x-15)3=53
4x-15=5
4x= 5+15
4x=20
vậy x = 20
Bài 2:
a: Ta có: x+11=26
nên x=15
b: Ta có: \(\left(4x-15\right)^3=125\)
\(\Leftrightarrow4x-15=5\)
hay x=5
a. 2x + 70 = 74
<=> 2x = 4
<=> x = 2
b. 120 - \(\dfrac{4x}{2}\) = 80
<=> 120 - 2x = 80
<=> 120 - 80 = 2x
<=> 2x = 40
<=> x = 20
c. (3x + 5)2 = 400
<=> \(|3x+5|=\sqrt{400}\)
<=> \(|3x+5|=20\)
<=> \(\left[{}\begin{matrix}3x+5=20\\3x+5=-20\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{-25}{3}\end{matrix}\right.\)
không có giá trị x vì 2125 không thể tách thành bình phương
\(a.\)
\(\left(x-19\right)\left(x+21\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-19=0\\x+21=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=19\\x=-21\end{matrix}\right.\)
\(S=\left\{19,-21\right\}\)
\(b.\)
\(\left(53-x\right)\left(41+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}53-x=0\\41+x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=53\\x=-41\end{matrix}\right.\)
\(S=\left\{53,-41\right\}\)
Tìm x:
a) (x-19).(x+21)=0
<=>\(\left\{{}\begin{matrix}x-19=0< =>x=19\\x+21=0< =>x=-21\end{matrix}\right.\)
Vậy pt trên có tập nghiệm là: S={19;-21}
b)(53-x).(41+x)=0
<=>\(\left\{{}\begin{matrix}53-x=0< =>x=53\\41+x=0< =>x=-41\end{matrix}\right.\)
Vậy pt trên có tập nghiệm là S={53;-41}
a) Ta có: \(\left|4-5x\right|=24\)
\(\Leftrightarrow\left[{}\begin{matrix}-5x+4=24\\-5x+4=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-5x=20\\-5x=-28\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{28}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{-4;\dfrac{28}{5}\right\}\)
b) Ta có: \(\left(8+x\right)\left(6-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=6\end{matrix}\right.\)
Vậy: \(x\in\left\{-8;6\right\}\)
a) \(\left|4-5x\right|=24\)
\(\Leftrightarrow\left[{}\begin{matrix}4-5x=24\\4-5x=-24\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{28}{5}\end{matrix}\right.\)
b) \(\left(8+x\right)\left(6-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}8+x=0\\6-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=6\end{matrix}\right.\)
a) 89 - (73 - x) = 20
73 - x = 89 - 20 = 69
x = 73 - 69 = 4
b) 140 : (x-8) = 7
x - 8 = 140 : 7 = 2
x = 8 + 2 = 10
\(a,89-\left(73-x\right)=20\\ 73-x=69\\ x=4\\ b,140:\left(x-8\right)=7\\ x-8=20\\ x=28\)
b) Ta có: \(x-43=\left(57-x\right)-50\)
\(\Leftrightarrow x-43=57-x-50\)
\(\Leftrightarrow x+x=7+43\)
\(\Leftrightarrow2x=50\)
hay x=25
Vậy: x=25
c) Ta có: (x-2)(8-x)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\8-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=8\end{matrix}\right.\)
Vậy: \(x\in\left\{2;8\right\}\)
(2125-125):x=100
2000:x=100
x=2000:100
x=20
a,
2125:X-125:X=100
=>[2125-125]:X=100
=>2000:X=100
=>X=2000:100
=>X=20
Vậy X=20