Tìm số nguyên x biết:
a) −435-4\dfrac{3}{5}−453 x 2432\dfrac{4}{3}234 ≤\le≤ x ≤\le≤ −235-2\dfrac{3}{5}−253 : 16151\dfrac{6}{15}1156
b) −413-4\dfrac{1}{3}−431 (12−16\dfrac{1}{2}-\dfrac{1}{6}21−61 ) ≤\le≤ x ≤23(13−12−37)\le\dfrac{2}{3}\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{7}\right)≤32(31−21−73)
c) x+135+x+234+x+333=x+432+x+531x+630\dfrac{x+1}{35}+\dfrac{x+2}{34}+\dfrac{x+3}{33}=\dfrac{x+4}{32}+\dfrac{x+5}{31}\dfrac{x+6}{30}35x+1+34x+2+33x+3=32x+4+31x+530x+6
c) x+135+x+234+x+333=x+432+x+531+x+630\dfrac{x+1}{35}+\dfrac{x+2}{34}+\dfrac{x+3}{33}=\dfrac{x+4}{32}+\dfrac{x+5}{31}+\dfrac{x+6}{30}35x+1+34x+2+33x+3=32x+4+31x+5+30x+6
⇒x+135+1+x+234+1+x+333+1=x+432+1+x+531+1+x+630+1\Rightarrow\dfrac{x+1}{35}+1+\dfrac{x+2}{34}+1+\dfrac{x+3}{33}+1=\dfrac{x+4}{32}+1+\dfrac{x+5}{31}+1+\dfrac{x+6}{30}+1⇒35x+1+1+34x+2+1+33x+3+1=32x+4+1+31x+5+1+30x+6+1
⇒x+1+3535+x+2+3434+x+3+3333=x+4+3232+x+5+3131+x+6+3030\Rightarrow\dfrac{x+1+35}{35}+\dfrac{x+2+34}{34}+\dfrac{x+3+33}{33}=\dfrac{x+4+32}{32}+\dfrac{x+5+31}{31}+\dfrac{x+6+30}{30}⇒35x+1+35+34x+2+34+33x+3+33=32x+4+32+31x+5+31+30x+6+30
⇒x+3635+x+3634+x+3633=x+3632+x+3631+x+3630\Rightarrow\dfrac{x+36}{35}+\dfrac{x+36}{34}+\dfrac{x+36}{33}=\dfrac{x+36}{32}+\dfrac{x+36}{31}+\dfrac{x+36}{30}⇒35x+36+34x+36+33x+36=32x+36+31x+36+30x+36
⇒x+3635+x+3634+x+3633−x+3632−x+3631−x+3630=0\Rightarrow\dfrac{x+36}{35}+\dfrac{x+36}{34}+\dfrac{x+36}{33}-\dfrac{x+36}{32}-\dfrac{x+36}{31}-\dfrac{x+36}{30}=0⇒35x+36+34x+36+33x+36−32x+36−31x+36−30x+36=0
⇒(x+36)(135+134+133+132+131+130)=0\Rightarrow\left(x+36\right)\left(\dfrac{1}{35}+\dfrac{1}{34}+\dfrac{1}{33}+\dfrac{1}{32}+\dfrac{1}{31}+\dfrac{1}{30}\right)=0⇒(x+36)(351+341+331+321+311+301)=0
⇒x+36=0(vıˋ 135+134+133+132+131+130e0)\Rightarrow x+36=0\left(\text{vì }\dfrac{1}{35}+\dfrac{1}{34}+\dfrac{1}{33}+\dfrac{1}{32}+\dfrac{1}{31}+\dfrac{1}{30}e0\right)⇒x+36=0(vıˋ 351+341+331+321+311+301e0)
⇒x=−36\Rightarrow x=-36⇒x=−36
Vậy ...
(4x+4)(3x+3)=32
Cho x,y,z > 0 thỏa xy+yz+zx = 1. tìm GTNN của x^2 + y^2 + z^2 B2: giải phương trình:
Cho tứ giác lồi ABCD có đường chéo AC bằng cạnh AD. C/minh: BC < BD
Cho tứ giác ABCD có AB//CD và AB=CD .Chứng minh BC=AD và BC//AC
a) Cho a^2 + b^2 + c^2 + 3 = 2(a+b+c). Chứng minh a=b=c=1
b) Cho (a+b+c)^2 = 3(ab+bc+ac). Chứng minh a+b+c
c) Cho (a+b)^2 + (b-c)^2 + (c-a)^2 = (a+b-2c^2) + (b+c-2a^2) + (c+a-2b)^2. Chứng minh a=b=c
Tìm x,biết:
a)64x3+48x2+12x+1=27
b)(x-5)2+9=0
giúp mk nhé.mk sẽ tick cho.cảm ơn trước!
Bài 20: Điền hàng tử thích hợp vào chỗ có dấu * để có hằng đẳng thức:
c) x2 + x + * = (* + *)2
d) * - 2a + 4 = (* - *)2
e) 4y2 - * = (* - 3x)(* + *)
f) * - 14\dfrac{1}{4}41= (3y - *)(* + *)
g) 8x3 + * = (* + 2a)(4x2 - * + *)
h) * - 27x3 = (4x - *)(9y2 + * + *)
i) * + * = (* + 12\dfrac{1}{2}21y)(4x2 - * + *)
j) * - * = (4y - *)(* + y + *)
k) 64a3 + * + * + 27b3b = (* + *)3
l) 8a3 - * + * - * = (* - 3y)3
m) (* - *)3 = x3 - * + 12xy2 - *
n) (* + *) = * + 108x2y + 144xy2 + *
Tìm x,y biết
x2 + y2 - 2x + 4y +5 =0
Tính giá trị biểu thức:
D= (2x - 3)2 - (4x - 6)(2x - 5) + (2x - 5)2 với x = 99
Tính giá trị của biểu thức:
A= 4x2 +8x+5 với x=49