BT6: Tìm x, biết a, x^2=25 b, x^2-1=15 c, 19-2x^2=1
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`@` `\text {Ans}`
`\downarrow`
`a,`
`x^2 + 2x + 1 = 9`
`=> x^2 + 2x + 1 - 9 = 0`
`=> x^2 + 2x - 8 = 0`
`=> x^2 + 4x - 2x - 8 = 0`
`=> (x^2 + 4x) - (2x + 8) = 0`
`=> x(x + 4) - 2(x + 4) = 0`
`=> (x-2)(x+4) = 0`
`=>`\(\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
Vậy, `x \in {2; 4}`
`b,`
`x^2 - 1 = 15`
`=> x^2 = 15 + 1`
`=> x^2 = 16`
`=> x^2 = (+-4)^2`
`=> x = +-4`
Vậy, `x \in {4; -4}`
`c)`
`19 - 2x^2 = 1`
`=> 2x^2 = 19 - 1`
`=> 2x^2 = 18`
`=> x^2 = 18 \div 2`
`=> x^2 = 9`
`=> x^2 = (+-3)^2`
`=> x = +-3`
Vậy, `x \in {3; -3}.`
1 Tìm số nguyên x bik:
a.-( 2x+2)+25=-19
-2x - 2 + 25 = - 19
-2x = -19+2-25
-2x = -42
x = 21
b.1-(12+3x)=7
1 - 12 - 3x = 7
- 11 - 3x = 7
-3x = 7 + 11
- 3x = 18
x = -6
c.-(2x+2)+25=-19
giống câu a nhé
2.Rút gọn biểu thức
a.2x+(-61)-(21-61)
= 2x - 61 - 21 + 61
= 2x - 21
b. (-3-x+5)+3
= 2 - x + 3
= -1 -x
c.11-(13-x)+(13-11)
= 11 - 13 + x + 13 - 11
= x
d.25-(15-x+303)+303
= 25 - 15 + x - 303 + 303
= 10 + x
e.x+(-81)-(11-8)
= x - 81 - 11 + 8
= x - 84
f. (-1-x+2)+1
= -1 - x + 2 + 1
= - x + 2
g.15-(11-x)+(11-15)
= 15 - 11 + x + 11 - 15
= x
i) 15-(15-x+202)+202
= 15 - 15 + x - 202 + 202
= x
3.Chứng minh đẳng thức
a.-(59-3x)+39=3x-20
- 59 + 3x + 39 = 3x - 20
- 59 + 39 + 20 = 3x - 3x
0 = 0
b.-(a+b+c)+(b-c)-(a-c-1)=1+c-29
b - a - b - c + b - c - a + c + 1 = 1 + c - 29
b - a - b - c + b - c - a + c - c = 1 - 29 - 1
- 2a + b - 2c = - 29
chịu luôn -__-
c.-(19-2x)+39=2x+20
- 19 + 2x + 39 = 2x + 20
2x - 2x = 20 + 19 - 29
0 = 0
d.-(a+b+c)+b-c-( a-c+1)=c-2a-1
- a - b - c + b - c - a + c - 1 = c - 2a - 1
- a - b - c + b - c - a + c + 2a - c = -1 + 1
- 2c = 0
c = 0
Vậy c = 0
a) 3x – 15 = 25 – 5x
=> 3x + 5x = 25 + 15
=> 8x = 40
=> x = 5
b) 3x - 17 = 2x – 7
=> 3x - 2x = -7 + 17
=> x = 10
c) 2x – 17 = – (3x – 18)
=> 2x - 17 = -3x + 18
=> 2x + 3x = 18 + 17
=> 5x = 35
=> x = 7
d) 3x – 14 = 2(x – 9) + 1
=> 3x - 14 = 2x - 18 + 1
=> 3x - 2x = -18 + 1 + 14
=> x = -3
f) (x – 5)2 = 9
\(\Rightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
a) Ta có: \(3x-15=25-5x\)
\(\Leftrightarrow3x-15-25+5x=0\)
\(\Leftrightarrow8x-40=0\)
\(\Leftrightarrow8x=40\)
hay x=5
Vậy: x=5
b) Ta có: \(3x-17=2x-7\)
\(\Leftrightarrow3x-17-2x+7=0\)
\(\Leftrightarrow x-10=0\)
hay x=10
Vậy: x=10
c) Ta có: \(2x-17=-\left(3x-18\right)\)
\(\Leftrightarrow2x-17=-3x+18\)
\(\Leftrightarrow2x-17+3x-18=0\)
\(\Leftrightarrow5x-35=0\)
\(\Leftrightarrow5x=35\)
hay x=7
Vậy: x=7
d) Ta có: \(3x-14=2\left(x-9\right)+1\)
\(\Leftrightarrow3x-14=2x-18+1\)
\(\Leftrightarrow3x-14-2x+18-1=0\)
\(\Leftrightarrow x+3=0\)
\(\Leftrightarrow x=-3\)
Vậy: x=-3
f) Ta có: \(\left(x-5\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{2;8\right\}\)
a) (5x+1) ^ 2 = 4^2 : 5^ 2
( 5x+1) ^2 = (4:5) ^2
=> (5x+1) = ( 4 : 5) = 0.8
5x = 0.8 - 1
x = 0.7 : 5
x = 0,14
Câu 1:
a: =>-2x-x+17=34+x-25
=>-3x+17=x+9
=>-4x=-8
hay x=2
b: =>17x+16x+27=2x+43
=>33x+27=2x+43
=>31x=16
hay x=16/31
c: =>-2x-3x+51=34+2x-50
=>-5x+51=2x-16
=>-7x=-67
hay x=67/7
e: 3x-32>-5x+1
=>8x>33
hay x>33/8
a) \(5\left(x-7\right)=0\)
\(\Rightarrow x-7=0\)
\(\Rightarrow x=7\)
b) \(25\left(x-4\right)=0\)
\(\Rightarrow x-4=0\)
\(\Rightarrow x=4\)
c) \(\left(34-2x\right)\left(2x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}34-2x=0\\2x-6=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=34\\2x=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=17\\x=3\end{matrix}\right.\)
d) \(\left(2019-x\right)\left(3x-12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\3x=12\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\x=\dfrac{12}{3}=4\end{matrix}\right.\)
e) \(57\left(9x-27\right)=0\)
\(\Rightarrow9x-27=0\)
\(\Rightarrow9\left(x-3\right)=0\)
\(\Rightarrow x-3=0\)
\(\Rightarrow x=3\)
a) 5.(x-7)=0⇔x-7=0⇔x=7
b) 25(x-4)=0⇔x-4=0⇔x=4
c) (34-2x).(2x-6)=0
⇔ 34-2x=0 hoặc 2x-6=0
⇔2x=34 hoặc 2x=6
⇔ x=17 hoặc x=3
d) (2019-x).(3x-12)=0
⇔ 2019-x=0 hoặc 3x-12=0
⇔ x=2019 hoặc x=4
e) 57.(9x-27)=0
⇔ 9x-27=0
⇔ x=3
f) 25+(15-x)=30
⇔ 15-x=5
⇔ x=10
g) 43-(24-x)=20
⇔ 24-x=23
⇔ x=1
h) 2.(x-5)-17=25
⇔ 2(x-5)=42
⇔x-5=21
⇔ x=26
i) 3(x+7)-15=27
⇔ 3(x+7)=42
⇔ x+7=14
⇔ x=7
j) 15+4(x-2)=95
⇔ 4(x-2)=80
⇔ x-2=20
⇔ x=22
k) 20-(x+14)=5
⇔ x+14=15
⇔ x=1
l) 14+3(5-x)=27
⇔ 3(5-x)=13
⇔ 5-x=13/3
⇔ x=5-13/3
⇔ x=2/3
a/
\(x^2=25\Leftrightarrow x=\pm5\)
b/
\(x^2-1=15\\\Leftrightarrow x^2=16\Leftrightarrow x=\pm4\)
c/
\(19-2x^2=1\Leftrightarrow2x^2=18\Leftrightarrow x^2=9\Leftrightarrow x=\pm3\)
`@` `\text {Ans}`
`\downarrow`
`a,`
`x^2 = 25`
`=> x^2 = (+-5)^2`
`=> x = +-5`
Vậy, `x \in {5; -5}`
`b,`
`x^2 - 1 = 15`
`=> x^2 = 15+1`
`=> x^2 = 16`
`=> x^2 = (+-4)^2`
`=> x = +-4`
Vậy, `x \in {4; -4}`
`c,`
`19 - 2x^2 = 1`
`=> 2x^2 = 19 - 1`
`=> 2x^2 = 18`
`=> x^2 = 18 \div 2`
`=> x^2 = 9`
`=> x^2 = (+-3)^2`
`=> x = +-3`
Vậy, `x \in {3; -3}.`