96 - 3(x + 1) = 42
Tìm x
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21 tháng 9 2021
2) \(A=-x^2-y^2+2x-6y+9=-\left(x^2-2x+1\right)-\left(y^2+6y+9\right)+19=-\left(x-1\right)^2-\left(y+3\right)^2+19\)
\(maxA=19\Leftrightarrow\)\(\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)
AD
4
16 tháng 12 2023
a. 96 : 8 : 2 = 6
96 : ( 8 x 2 ) = 6
96 : 8 : 2 = 96 : ( 8 x 2 )
b. 81 : 9 : 3 = 3
81 : ( 9 x 3 ) = 3
81 : 9 : 3 = 81 : ( 9x 3)
N
1
28 tháng 2 2022
a: \(\dfrac{96}{\left(x-4\right)\left(x+4\right)}+\dfrac{7+x}{4-x}=\dfrac{2x-1}{x+4}-3\)
\(\Leftrightarrow\dfrac{96}{\left(x-4\right)\left(x+4\right)}-\dfrac{\left(x+7\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{\left(2x-1\right)\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}-\dfrac{3\left(x-4\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}\)
Suy ra: \(96-x^2-11x-28=2x^2-9x+4-3\left(x^2-16\right)\)
\(\Leftrightarrow-x^2-11x+68=2x^2-9x+4-3x^2+48\)
\(\Leftrightarrow-x^2-11x+68=-x^2-9x+52\)
=>-11x+68=-9x+52
=>-2x=-16
hay x=8(nhận)
b: \(\dfrac{2}{x-1}+\dfrac{3}{x-2}=\dfrac{3}{x-3}\)
\(\Leftrightarrow2\left(x-2\right)\left(x-3\right)+3\left(x-1\right)\left(x-3\right)=3\left(x-1\right)\left(x-2\right)\)
\(\Leftrightarrow2\left(x^2-5x+6\right)+3\left(x^2-4x+3\right)=3\left(x^2-3x+2\right)\)
\(\Leftrightarrow2x^2-10x+12+3x^2-12x+9=3x^2-9x+6\)
\(\Leftrightarrow5x^2-22x+21-3x^2+9x-6=0\)
\(\Leftrightarrow2x^2-13x+15=0\)
\(\Leftrightarrow2x^2-10x-3x+15=0\)
=>(x-5)(2x-3)=0
=>x=5(nhận) hoặc x=3/2(nhận)
6 tháng 3 2020
\(\frac{x+2}{98}+1+\frac{x+3}{97}+1=\frac{x+4}{96}+1+\frac{x+5}{95}+1\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{96}+\frac{x+100}{95}\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{97}-\frac{x+100}{96}-\frac{x+100}{95}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)
\(\Leftrightarrow x+100=0\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\ne0\right)\)
<=> x=-100
6 tháng 3 2020
ko chép đề nhé
\(\frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{96}+\frac{x+100}{95} \)
=> \((x+100)(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95})=0\)
vì \((\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}) khác 0\)
=>\(x+100=0\)
=>x=-100
16 tháng 5 2022
\(\Leftrightarrow\dfrac{x+2+98}{98}+\dfrac{x+3+97}{97}=\dfrac{x+4+96}{96}+\dfrac{x+5+95}{95}\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\right)=0\)
=>x+100=0
hay x=-100
LV
19 tháng 8 2018
\(x+x\times\frac{1}{3}=\frac{96}{108}\)
\(x\times\left(1+\frac{1}{3}\right)=\frac{8}{9}\)
\(x\times\frac{4}{3}=\frac{8}{9}\)
\(x=\frac{8}{9}\div\frac{4}{3}\)
\(x=\frac{2}{3}\)
Vậy .............
NM
19 tháng 8 2018
Ta có : \(x\)+ \(x\)x \(\frac{1}{3}\)= \(\frac{96}{108}\)
\(x\)x \(\left(1+\frac{1}{3}\right)\)= \(\frac{96}{108}\)
\(x\)x \(\frac{4}{3}\)= \(\frac{96}{108}\)
\(x\) = \(\frac{96}{108}\): \(\frac{4}{3}\)
\(x\) = \(\frac{96}{108}\)x \(\frac{3}{4}\)
\(x\) = \(\frac{24}{27}\)x \(\frac{3}{4}\)
\(x\) = \(\frac{6}{9}\)
C
17 tháng 3 2023
`(x+1)/99+(x+2)/98+(x+3)/97+(x+4)/96=-4`
`=>(x+1)/99+1+(x+2)/98+1+(x+3)/97+1+(x+4)/96+1=-4+4`
`=>(x+100)/99+(x+100)/98+(x+100)/97+(x+100)/96=0`
`=>(x+100)(1/99+1/98+1/97+1/96)=0`
`=>x+100=0` (Vì `1/99+1/98+1/97+1/96\ne0`)
`=>x=-100`
Vậy ...
`#`𝐷𝑎𝑖𝑙𝑧𝑖𝑒𝑙
17 tháng 3 2023
\(\dfrac{x+1}{99}+\dfrac{x+2}{98}+\dfrac{x+3}{97}+\dfrac{x+4}{96}=-4\\ \dfrac{x+1}{99}+\dfrac{x+2}{98}+\dfrac{x+3}{97}+\dfrac{x+4}{96}+4=0\\ \left(\dfrac{x+1}{99}+1\right)+\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)+\left(\dfrac{x+4}{96}+1\right)=0\\ \dfrac{x+100}{99}+\dfrac{x+100}{98}+\dfrac{x+100}{97}+\dfrac{x+100}{96}=0\\ \left(x+100\right)\left(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+\dfrac{1}{96}\right)=0\)
mà `1/99+1/98+1/97+1/96 \ne 0`
nên `x+100=0`
`x=-100`
NQ
2
2 tháng 10 2021
a: Ta có: \(\left(\dfrac{3}{2}x-\dfrac{1}{5}\right)^2\cdot\left(x^2+\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow x\cdot\dfrac{3}{2}=\dfrac{1}{5}\)
hay \(x=\dfrac{1}{5}:\dfrac{3}{2}=\dfrac{2}{15}\)
b: Ta có: \(\dfrac{x+1}{99}+\dfrac{x+2}{98}+\dfrac{x+3}{97}+\dfrac{x+4}{96}=-4\)
\(\Leftrightarrow x+100=0\)
hay x=-100
6 tháng 8 2021
a, \(15:\left(x+2\right)=3\Leftrightarrow x+2=3\Leftrightarrow x=1\)
b, \(20:\left(x+1\right)=2\Leftrightarrow x+1=10\Leftrightarrow x=9\)
c, \(240:\left(x-5\right)=2^2.5^2-20=80\Leftrightarrow x-5=3\Leftrightarrow x=8\)
d, \(96-3\left(x+1\right)=42\Leftrightarrow3\left(x+1\right)=54\Leftrightarrow x+1=18\Leftrightarrow x=17\)
6 tháng 8 2021
a) 15 : (x + 2) = 3
x + 2 = 15 : 3
x + 2 = 5
x = 5 – 2 = 3
b) 20 : (1 + x) = 2
1 + x = 20 : 2
1 + x = 10
x = 10 – 1 = 9
c) 240 : (x – 5) = 22.52 – 20
240 : (x – 5) = 4.25 – 20
240 : (x - 5) = 100 – 20
240 : (x - 5) = 80
x – 5 = 240 : 80
x – 5 = 3
x = 3 + 5 = 8
d) 96 - 3(x + 1) = 42
3(x + 1) = 96 – 42
3(x + 1) = 54
x + 1 = 54 : 3
x + 1 = 18
x = 18 – 1
x = 17
=> 3(x+1)=96-42
=> x+1=54:3
=> x=18-1
=> x=17
X=17 k đi