Tìm x :
X + 10 = 20
X - 20 = 30
X . 20 = 40
X : 30 = 60
Kb nha
Cho . Tìm ; biết .
; . ; . ; . ; .Lời giải:
$3x=16y\Rightarrow \frac{x}{16}=\frac{y}{3}$
Áp dụng TCDTSBN:
$\frac{x}{16}=\frac{y}{3}=\frac{x+y}{16+3}=\frac{190}{19}=10$
$\Rightarrow x=10.16=160; y=3.10=30$
Đáp án A.
Tìm x
1500 : [ ( 30x + 40x ) : x ] = 30
x +20x x+10x x+40x x+30=205000 giải giúp mình nhé
\(x+20\times x+10\times x+40\times x+30=205000\)
\(x\times\left(1+20+10+40\right)+30=205000\)
\(x\times71+30=205000\)
\(x\times71=205000-30\)
\(x\times71=204970\)
\(x=204970:71\)
\(x=\frac{204970}{71}\)
tìm x:
8.(x-6)+4.(x-3)-2.(+7)=-x+4
-7.(8-x)-6.(9+x)=20-x
9.(x-7)-20.(4-3x)=-7x+15
-16.(6+x)-30.(2-x)=-40x-16
17.(x-6)-14.(x+2)=4.(x-6)-2.(x-2)
mn giải giúp mình vs ạ.Mình đang cần gấp!!!!
\(8x-48+4x-12-14=-x+4\)
\(\Leftrightarrow12x-75=-x+4\Leftrightarrow13x=79\Leftrightarrow x=\dfrac{79}{13}\)
\(-7\left(8-x\right)-6\left(x+9\right)=20-x\Leftrightarrow-56+7x-6x-54=20-x\)
\(\Leftrightarrow2x=130\Leftrightarrow x=65\)
\(9x-63-80+60x=-7x+15\Leftrightarrow76x=158\Leftrightarrow x=\dfrac{79}{38}\)
\(-96-16x-60+30x=-40x-16\Leftrightarrow54x=140\Leftrightarrow x=\dfrac{70}{27}\)
\(17x-102-14x-28=4x-24-2x+4\Leftrightarrow x=110\)
x +20x x+10xx+40x x+30=205000 bằng mấy vậy mình đang cực gấp luôn mong các bn sẽ giải giúp mình mình sẽ tick cho
cho đa thức fx= x^2020-20.x^2019+20.x^2018-...-20x+30 tính f(19)
Tìm x :
16x + 20x + 24x + .............. + 40x + 44x = 4800
16x+20x+24x+................+40x+44x=4800
=(16+20+24+.....+40+44)x=4800
=240x=4800
x=20
16x + 20x + 24x + .............. + 40x + 44x = 4800
(16 + 20 + 24 + ..................+ 40 + 44). x = 4800
\(\frac{\left(16+44\right).8}{2}\). x = 4800
240 . x = 4800
x = 4800 : 240
x = 20
đáp số: 20
16x+20x+24x+........+40x+44x=4800
(16+20+.......+40+44)x=4800
240x=4800
x=4800:240
x=20
tìm x biết :
1) x(x-5)-4x+20=0
2) x(x+6)-7x-42=0
3) x4-2x3+10x2-20x=0
4) x2+3x-18=0
5) 8x2+3x+7=0
6) x3-11x2+30x=0
Bài 1:Tính giá trị nhỏ nhất của biểu thức:
A=X^2-20x+101
B=2x^2+40x-1
C=x^2-4xy+5y^2-2y+28
D=(x-2).(x-5).(x^2-7x-10)
\(A=x^2-20x+101=\left(x-10\right)^2+1\ge1\)
\(minA=1\Leftrightarrow x=10\)
\(B=2x^2+40x-1=2\left(x+10\right)^2-201\ge-201\)
\(minB=-201\Leftrightarrow x=-10\)
\(C=x^2-4xy+5y^2-2y+28=\left(x^2-4xy+4y^2\right)+\left(y^2-2y+1\right)+27=\left(x-2y\right)^2+\left(y-1\right)^2+27\ge27\)
\(minC=27\Leftrightarrow\)\(\left\{{}\begin{matrix}y=1\\x=2\end{matrix}\right.\)
\(D=\left(x-2\right)\left(x-5\right)\left(x^2-7x-10\right)=\left(x^2-7x+10\right)\left(x^2-7x+10\right)=\left(x^2-7x\right)^2-100\ge-100\)
\(minD=100\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\x=7\end{matrix}\right.\)
a: Ta có: \(A=x^2-20x+101\)
\(=x^2-20x+100+1\)
\(=\left(x-10\right)^2+1\ge1\forall x\)
Dấu '=' xảy ra khi x=10
b: ta có: \(B=2x^2+40x-1\)
\(=2\left(x^2+20x-\dfrac{1}{2}\right)\)
\(=2\left(x^2+20x+100-\dfrac{201}{2}\right)\)
\(=2\left(x+10\right)^2-201\ge-201\forall x\)
Dấu '=' xảy ra khi x=-10