Tìm x biết:
(x + 1)(x + 2)(x + 3)(x + 4) = 24
Tìm x, biết: a) x = 1/4 + 5/13 b) x/3 = 2/3 + -1/7 c) x/3 = 16/24 + 24/ 36
d) x/15 = 1/5 + 2/3
\(a)x=\dfrac{1}{4}+\dfrac{5}{13}=\dfrac{33}{52}.\\ b)\dfrac{x}{3}=\dfrac{2}{3}+\dfrac{-1}{7}.\\ \Leftrightarrow\dfrac{x}{3}=\dfrac{11}{21}.\\ \Leftrightarrow\dfrac{7x}{21}=\dfrac{11}{21}.\\ \Rightarrow7x=11.\\ \Leftrightarrow x=\dfrac{11}{7}.\\ c)\dfrac{x}{3}=\dfrac{16}{24}+\dfrac{24}{36}=\dfrac{2}{3}+\dfrac{2}{3}=\dfrac{4}{3}.\\ \Rightarrow x=4.\\ d)\dfrac{x}{15}=\dfrac{1}{5}+\dfrac{2}{3}=\dfrac{13}{15}.\\ \Rightarrow x=13.\)
tìm x biết (x+1)(x+2)(x+3)(x+4)-24=0
Tìm x biết: a) \(\dfrac{6}{-x}=\dfrac{x}{-24}\) b) \(x-\dfrac{7}{12}x+\dfrac{3}{8}x=\dfrac{5}{24}\)
c)\(\left(x-\dfrac{1}{3}\right)^2-\dfrac{1}{2}=1\dfrac{3}{4}\) d) \(\dfrac{x-3}{-2}=\dfrac{-8}{x-3}\)
e) \(\dfrac{9}{x}=\dfrac{-35}{105}\) f) \(\left(x-\dfrac{1}{2}\right)\left(-3-\dfrac{x}{2}\right)=0\)
a: =>6/x=x/24
=>x^2=144
=>x=12 hoặc x=-12
b: =>x(1-7/12+3/8)=5/24
=>x*19/24=5/24
=>x=5/24:19/24=5/19
c: =>(x-1/3)^2=1+3/4+1/2=9/4
=>x-1/3=3/2 hoặc x-1/3=-3/2
=>x=11/6 hoặc x=-7/6
d: =>(x-3)^2=16
=>x-3=4 hoặc x-3=-4
=>x=-1 hoặc x=7
e: =>9/x=-1/3
=>x=-27
f: =>x-1/2=0 hoặc -x/2-3=0
=>x=1/2 hoặc x=-6
tìm x biết
a, (x^2+x)+4(x^2+x)-12=0
b, (x+1)(x+2)(x+3)(x+4)-24=0
Tìm x biết
4(2x+7) - 3(3x - 2)=24
-2(x+3)+(-4)^2=3(1-x)
a) 4(2x+7)-3(3x-2)=24
8x+28-9x-6=24
(8x-9x)+(28-6)=24
(-1)x+22=24
(-1)x=24-22=2
x=2:(-1)=-2
b) -2(x+3)+(-4)2=3(1-x)
(-2)x+(-6)+16=3-3x
(-2)x+(-6)+16+3x=3
[(-2)x+3x]+[(-6)+16]=3
x+10=3
x=3-10=-7
Tìm x ϵ Z, biết:
1/2 - (1/3 + 3/4) ≤ x ≤ 1/24 - (1/8 - 1/3)
`1/2-(1/3+3/4)<=x<=1/24-(1/8-1/3)`
`<=>6/12-4/12-9/12<=x<=1/24-3/24+8/24`
`<=>-7/12<=x<=1/4`
`<=>-14/24<=x<=3/12`
`=>-14<=x<=3`
`=>x\in{-14;-13;-12;...;3}` do `x\inZZ`
a) Tìm x biết (x + 3)2 = (x + 3)(x – 3)
b) Chứng tỏ A = (x + 1)(x +2)(x + 3)(x + 4) – 24 chia hết cho x (với x ≠ 0)
Ta có:\(\left(x+3\right)^2=\left(x+3\right)\left(x-3\right)\)
Xét \(x+3=0\Rightarrow x=-3\)
Xét \(x+3\ne0\) ta có:
\(x+3=x-3\)
\(\Rightarrow0=6\left(VL\right)\)
Vậy \(x=-3\)
a)
(x + 3)2 = (x + 3)(x – 3)
⇔ (x + 3)2 - (x + 3)(x - 3) = 0
⇔ (x + 3)(x + 3 - x + 3) = 0
⇔ 6(x + 3) = 0
⇔ x = -3
Vậy: x = -3
b) Ta có A = (x + 1)(x + 2)(x + 3)(x + 4) – 24
= (x + 1)(x + 4)(x + 2)(x + 3) - 24
= (x2 + 5x + 4)(x2 + 5x + 6) - 24(*)
Đặt x2 + 5x + 5 = t
Thay x2 + 5x + 5 = t vào (*) ta được:
A = (t - 1)(t + 1) - 24
= t2 - 25
= (t - 25)(t + 25)
= (x2 + 5x + 5 + 5)(x2 + 5x + 5 - 5)
= (x2 + 5x + 10)(x2 + 5x)
(x2 + 5x + 10).x(x + 5) chia hết cho x (Với x ≠ 0)
Vậy: A chia hết cho x (Với x ≠ 0)
a) (x + 3)2 = (x + 3)(x - 3)
<=> x2 + 6x + 9 = x2 - 32
<=> x2 + 6x + 9 = x2 - 9
<=> 6x + 9 = -9
<=> 6x = -9 - 9
<=> 6x = -18
<=> x = -3
=> x = -3
1 : Tìm x , biết :
a, 4x - 5 . ( -3 + x ) = 7
b, -4 . ( x + 1 ) + 8 . ( x - 3 ) = 24
c, 7. ( 5 - x ) + 5 . ( x - 2 ) = 15
d, 4 . ( x - 1 ) - 3 . ( x - 2 ) = -5
Tìm B biết \(B=\frac{1+x^2+x^4+...+x^{22}+x^{24}+x^{26}}{1+x^4+x^8+...+x^{16}+x^{20}+x^{24}}\)
\(B=\frac{1+x^2+x^4+...+x^{26}}{1+x^4+x^8+...+x^{24}}\)
\(=\frac{\frac{\left(x^2-1\right)\left(1+x^2+x^4+...+x^{26}\right)}{x^2-1}}{\frac{\left(x^4-1\right)\left(1+x^4+x^8+...+x^{24}\right)}{x^4-1}}\)
\(=\frac{\frac{x^{28}-1}{x^2-1}}{\frac{x^{28}-1}{x^4-1}}=\frac{x^4-1}{x^2-1}=x^2+1\)