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a) \(\left(2,4-3x\right).0,5=0,9\)
\(2,4-3x=0,9:0,5\)
\(2,4-3x=1,8\)
\(3x=2,4-1,8\)
\(3x=0,6\)
\(x=0,6:3\)
\(x=0,2\)
b) \(\left(8,8x-50\right):0,4=51\)
\(8,8x-50=51.0,4\)
\(8,8x-50=20,4\)
\(8,8x=20,4+50\)
\(8,8x=70,4\)
\(x=70,4:8,8\)
\(x=8\)
c) \(\left(0,472-x\right):2=1,634\)
\(0,472-x=1,634.2\)
\(0,472-x=3,268\)
\(x=0,472-3,268\)
\(x=-2,796\)
d) \(1,6-\left(x-0,2\right)=6,5\)
\(x-0,2=1,6-6,5\)
\(x-0,2=-4,9\)
\(x=-4,9+0,2\)
\(x=-4.7\)
\(a.2,4-3x=1,8\)
\(3x=0,6\)
\(x=0,2\)
\(b.8,8x-50=20,4\)
\(8,8x=70,4\)
\(x=8\)
\(c.0,472-x=3,268\)
\(x=-2,796\)
\(d.x-0,2=-4,9\)
\(x=-4,7\)
a: \(\left(x+10\right)\left(x-5\right)=0\)
=>\(\left[{}\begin{matrix}x+10=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-10\\x=5\end{matrix}\right.\)
b: \(\left(2x+10\right)\left(4+x\right)=0\)
=>\(\left[{}\begin{matrix}2x+10=0\\4+x=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-4\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-5\end{matrix}\right.\)
c: \(\left(4x+20\right)\left(12x-24\right)=0\)
=>\(\left[{}\begin{matrix}4x+20=0\\12x-24=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-20\\12x=24\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
d: \(\left(x-2024\right)\left(4x+4\right)=0\)
=>\(\left[{}\begin{matrix}x-2024=0\\4x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2024\\4x=-4\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-1\\x=2024\end{matrix}\right.\)
e: \(\left(2x-6\right)\left(7+x\right)=0\)
=>\(\left[{}\begin{matrix}2x-6=0\\x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\x=-7\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=3\\x=-7\end{matrix}\right.\)
g: (4x+8)(6-x)=0
=>\(\left[{}\begin{matrix}4x+8=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x=6\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-2\\x=6\end{matrix}\right.\)
h: (2x+2)(4x-8)=0
=>2(x+1)*4*(x-2)=0
=>(x+1)(x-2)=0
=>\(\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
i: (2x-2024)(8x-16)=0
=>\(2\left(x-1012\right)\cdot8\cdot\left(x-2\right)=0\)
=>\(\left(x-1012\right)\left(x-2\right)=0\)
=>\(\left[{}\begin{matrix}x-1012=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1012\\x=2\end{matrix}\right.\)
a) 5 + 45(2x - 1) = 10
45(2x - 1) = 10 - 5
45(2x - 1) = 5
2x - 1 = 5 : 45
2x - 1 = 1/9
2x = 1/9 + 1
2x = 10/9
x = 10/9 : 2
x = 5/9
b) 54 : (2ˣ⁻³ + 1) + 3 = 9
54 : (2ˣ⁻³ + 1) = 9 - 3
54 : (2ˣ⁻³ + 1) = 6
2ˣ⁻³ + 1 = 54 : 6
2ˣ⁻³ + 1 = 9
2ˣ⁻³ = 9 - 1
2ˣ⁻³ = 8
2ˣ⁻³ = 2³
x - 3 = 3
x = 3 + 3
x = 6
c) 14 + 36 : 3ˣ⁻⁵ = 18
36 : 3ˣ⁻⁵ = 18 - 14
36 : 3ˣ⁻⁵ = 4
3ˣ⁻⁵ = 36 : 4
3ˣ⁻⁵ = 9
3ˣ⁻⁵ = 3²
x - 5 = 2
x = 2 + 5
x = 7
a: =>45(2x-1)=5
=>2x-1=1/9
=>2x=10/9
=>x=5/9
b: =>\(\dfrac{54}{2^{x-3}+1}=6\)
=>\(2^{x-3}+1=9\)
=>\(2^{x-3}=8\)
=>x-3=3
=>x=6
c: \(14+36:3^{x-5}=18\)
=>\(\dfrac{36}{3^{x-5}}=18-14=4\)
=>\(3^{x-5}=9\)
=>x-5=2
=>x=7
a) 3x-5 ⋮ x+2
+ (x+2) ⋮ (x+2)
⇒ 3(x+2) ⋮ (x+2)
⇒3x+6 ⋮ x+2
mà 3x-5 ⋮ x+2
⇒ 3x-5-(3x+6) ⋮ x+2
⇒ 3x-5-3x-6 ⋮ x+2
⇒ 3x-3x-5-6 ⋮ x+2
⇒-1 ⋮ x+2
⇒ x+2=-1
x =-1+2
x =1
vậy x=1
*câu b bnj cho đề bài rõ ràng hơn nhé
nếu đúng thì tích đúng cho mình nha
( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ... + ( x + 100 ) = 5750
( x + x + ... + x ) + ( 1 + 2 + ... + 100 ) = 5750
SSH là : ( 100 - 1) : 1 + 1 = 100 ( số )
Tổng là : ( 100 + 1 ) . 100 : 2 = 5050
=> 100x + 5050 = 5750
=> 100x = 700
=> x = 7
Vậy x = 7
\(\frac{7}{x}=\frac{y}{27}=\frac{-42}{54}\)
+) => \(\frac{y\cdot2}{27\cdot2}=\frac{-42}{54}\)
=> \(y\cdot2=-42\)
\(\Rightarrow y=-21\)
+) \(\frac{7}{x}=\frac{-21}{27}\)
=> \(\frac{7\cdot\left(-3\right)}{x\cdot\left(-3\right)}=\frac{-21}{27}\)
=> \(x\cdot\left(-3\right)=27\)
=> \(x=-9\)
* Lớp 6 chưa học đến tỉ lệ thức nên đây là cách đơn giản nhất r *
\(\frac{y}{27}\)=\(\frac{-42}{54}\)=>\(\frac{2y}{54}\)=\(\frac{-42}{54}\)=>2y= -42 y= \(\frac{-42}{2}\)= -21 \(\frac{7}{x}\)=\(\frac{-21}{27}\)=> -21x=7*27=189 x=\(\frac{189}{-21}\)= -9 Vậy x= -9, y= -21
a: x:1,25=-2,4
=>\(x=-2,4\cdot1,25=-3\)
b: \(\dfrac{5}{8}x-\dfrac{7}{24}=-\dfrac{2}{3}\)
=>\(\dfrac{5}{8}x=-\dfrac{2}{3}+\dfrac{7}{24}=-\dfrac{16}{24}+\dfrac{7}{24}=-\dfrac{9}{24}=-\dfrac{3}{8}\)
=>\(x=-\dfrac{3}{8}:\dfrac{5}{8}=-\dfrac{3}{8}\cdot\dfrac{8}{5}=-\dfrac{3}{5}\)