20 - 1 x 2/4 + 10 = ?
Tính (theo mẫu):
Mẫu: 5 x (10 – 4)
Cách 1:
5 x (10 – 4) = 5 x 6 = 30
Cách 2:
5 x (10 – 4) = 5 x 10 – 5 x 4 = 50 – 20 = 30
a) 3 x (20 – 5)
b) 20 x (40 – 1)
Hướng dẫn giải:
a) 3 x (20 – 5)
Cách 1:
3 x (20 – 5) = 3 x 15 = 45
Cách 2:
3 x (20 – 5) = 3 x 20 – 3 x 5 = 60 – 15 = 45
b) 20 x (40 – 1)
Cách 1:
20 x (40 – 1) = 20 x 39 = 780
Cách 2:
20 x (40 – 1) = 20 x 40 – 20 x 1 = 800 – 20 = 780
1. tìm x
a)1/2+X=5/6 b)X+1/4=3/4
c) 3/10 +X=1/2 d) X+1/4=3/8
2. tính giá trị biểu thức
7/20 -(5/8-2/5) 9/10 - (2/5+3/10)+7/20
Bài 1:
a: x+1/2=5/6
nên x=5/6-1/2=1/3
b: x+1/4=3/4
nên x=3/4-1/4=2/4=1/2
c: x+3/10=1/2
nên x=1/2-3/10=5/10-3/10=1/5
d: x+1/4=3/8
nên x=3/8-1/4=3/8-2/8=1/8
Bài 1: Tìm x
a. (-20) + x = -30 b. (-10) - x = -20
c. -10 + (-2) = -4 d. x + (-3) = -7
e. x - (-5) = -9 f. x (-11) = 12
h. 2x - 10 = 20 l. 4x - 8 = -8
k. -12 - (-2)x = -8
Bài 2: Tìm x
a. -20 - (10-x) = -3
b. 14 + (14-x) = -2
c. -15 - (x-3) = -7
d. (x+4) + (-20) = -8
e. -2x - 2 = -4
f. -2x + 4 = -4
l. -12 - (-2)x = -2 -4
Thank mn ạaa!!
Bài 1:
a. $(-20)+x=-30$
$x-20=-30$
$x=-30+20=-(30-20)=-10$
b.
$(-10)-x=-20$
$x=(-10)-(-20)=-10+20=20-10=10$
c. Đề sai. Bạn xem lại.
d.
$x+(-3)=-7$
$x=-7-(-3)=-7+3=-(7-3)=-4$
e.
$x-(-5)=-9$
$x=(-9)+(-5)=-14$
f.
$x(-11)=12$
$x=\frac{12}{-11}=\frac{-12}{11}$
h.
$2x-10=20$
$2x=20+10=30$
$x=30:2=15$
l.
$4x-8=-8$
$4x=-8+8=0$
$x=0:4=0$
k.
$-12-(-2)x=-8$
$(-2)x=-12-(-8)=-12+8=-(12-8)=-4$
$x=(-4):(-2)=2$
Bài 2:
a. $-20-(10-x)=-3$
$10-x=-20-(-3)=-20+3=-(20-3)=-17$
$x=10-(-17)=10+17=27$
b.
$14+(14-x)=-2$
$14-x=-2-14=-16$
$x=14-(-16)=14+16=30$
c.
$-15-(x-3)=-7$
$x-3=-15-(-7)=-15+7=-8$
x=-8+3=-5$
d.
$(x+4)+(-20)=-8$
$x+4=-8-(-20)=-8+20=12$
$x=12-4=8$
e.
$-2x-2=-4$
$-2x=-4+2=-2$
$x=(-2):(-2)=1$
f.
$-2x+4=-4$
$-2x=-4-4=-8$
$x=(-8):(-2)=4$
l.
$-12-(-2)x=-2-4=-6$
$(-2)x=-12-(-6)=-12+6=-6$
$x=(-6):(-2)=3$
20 x 1 + 20 x 2 + 20 x 3 + 20 x 4 + 20 x 5 + 20 x 6 + 20 x 7 + 20 x 8 + 20 x 9 + 20 x 10 =
A=2 x 6 x 10 + 4 x12 x 20 +6 x 18 x 30 + ...+ 20 x 60 x 100 / 1 x 2 x 3 + 2 x 4 x 6 + 3 x 6 x 9 +...+ 10 x 20 x 30
Khó lắm đấy . Ai giải được bái luôn
1 x 3 x 5 + 2 x6 x10 + 4 x 12 x 20
1 x 5 x 7 + 2 x 10 x 14 + 4 x 20 x 28
1 x 3 x 5 + 2 x 6 x 10 + 4 x 12 x 20
=15 + 120 + 960
= 135 + 960
=1095
1 x5 x 7 + 2 x 10 x 14 + 4 x 20 x 28
=35 + 280 + 2240
= 2555
k nha
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So sánh 2 phân số
1 x 3 x 5 +2 x 6 x 10 + 4 x 12 x 20 /1 x 5 x 7 + 2 x 10 x 14 + 4 x 20 x 28 và 3/8
Nhanh nhé !!! Mình đang cần gấp !!!
Tìm x, biết:
a) 3/(x+2)(x+5) + 5/(x+5)(x+10) + 7/ (x+10)(x+7) = x/(x+2)(x+17) với x thuộc {-2,-5,-10,-17}
b) 2/(x-1)(x-3 + 5/(x-3)(x-8) + 12/(x-8)(x-20) -1/x-20 =-3/4 với x thuộc {1,3,8,,20}
[1/(x+2)-1/(x+5)]+[1/(x+5)-1/(x+10)]+[1/(x+10)-1/(x+17)]=x/15.[1/(x+2)-1/(x+17)]
1/(x+2)-1/(x+17)=x/15.[1/(x+2)-1/(x+17)]
1=x/15
x=15
tìm x,y biết:
a) \(x^2+\left(y-\dfrac{1}{10}\right)^4=0\)
b) \(\left(\dfrac{1}{2}.x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\)
a) \(x^2+\left(y-\dfrac{1}{10}\right)^4=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y-\dfrac{1}{10}=0\end{matrix}\right.\)( do \(x^2\ge0,\left(y-\dfrac{1}{10}\right)^4\ge0\))
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{1}{10}\end{matrix}\right.\)
b) \(\left(\dfrac{1}{2}.x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x-5=0\\y^2-\dfrac{1}{4}=0\end{matrix}\right.\)( do \(\left(\dfrac{1}{2}x-5\right)^{20}\ge0,\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\))
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x=5\\y^2=\dfrac{1}{4}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=\pm\dfrac{1}{2}\end{matrix}\right.\)
\(a,\Leftrightarrow\left\{{}\begin{matrix}x=0\\y-\dfrac{1}{10}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{1}{10}\end{matrix}\right.\\ b,\left\{{}\begin{matrix}\left(\dfrac{1}{2}x-5\right)^{20}\ge0\\\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\end{matrix}\right.\Leftrightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\)
Mà \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\)
\(\Leftrightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}=0\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x=5\\y^2=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=\pm\dfrac{1}{2}\end{matrix}\right.\)
a) \(x^2+\left(y-\dfrac{1}{10}\right)^4=0\)
Mà \(x^2+\left(y-\dfrac{1}{10}\right)^4\ge0\forall x;y\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=0\\\left(y-\dfrac{1}{10}\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{1}{10}\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(0;\dfrac{1}{10}\right)\)
b) \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\)
Mà \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\forall x;y\)
\(\Rightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}\left(\dfrac{1}{2}x-5\right)^{20}=0\\\left(y^2-\dfrac{1}{4}\right)^{10}=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=10\\\left[{}\begin{matrix}y=\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)
Vậy \(\left(x;y\right)\in\left\{\left(10;\dfrac{1}{2}\right);\left(10;-\dfrac{1}{2}\right)\right\}\)
Tìm x, biết:
a) 3/(x+2)(x+5) + 5/(x+5)(x+10) + 7/ (x+10)(x+7) = x/(x+2)(x+17) với x thuộc {-2,-5,-10,-17}
b) 2/(x-1)(x-3 + 5/(x-3)(x-8) + 12/(x-8)(x-20) -1/x-20 =-3/4 với x thuộc {1,3,8,,20}
HELP ME!!!!
a: \(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
=>\(\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
=>x=15
b: \(\Leftrightarrow-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-8}-\dfrac{1}{x-8}+\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-3}{4}\)
=>1/x-1=3/4
=>x-1=4/3
=>x=7/3