b 120 : x - 2; 240 : x - 2 va x > c x + 2 :18 va 180 < x < 300 D x : 24 ; x : 38 va 100 < x < 320
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a) x - 120: 30 = 40
x -40 =40
x =40+40
x =80
b) (x + 120) : 20 = 8
(x+ 120) = 8x20
x+120 =160
x = 160-120
x = 40
c) (x + 5). 3 = 300
x+5=300:3
x+5=100
x=100-5
x=95
d) x.2 + 21 : 3= 27
x.2 +7=27
x.2 = 27-7
x.2= 20
x=20:2
x=10
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a. ( x + 3 )( x - 3 ) = 16
⇔x2-9=16
⇔x2-16-9=0
⇔x2-25=0
⇔(x-5)(x+5)=0
⇔\(\left[{}\begin{matrix}x+5=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=5\end{matrix}\right.\)
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A, 120 x 2 = 120 x 4
120 x 2 = 240
B, số đó là :
(2270 - 1034) : 4 = 309
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\(\sqrt{\left(120-11\right)^2}+\sqrt{\left(10-\sqrt{120}\right)^2}\)
\(=120-11+10+\sqrt{120}\)
\(=\sqrt{120}\left(\sqrt{120}+1\right)-1\)
\(a,=\left(120-11\right)+\left|10-\sqrt{120}\right|=109+\sqrt{120}-10=99+2\sqrt{30}\\ b,=\sqrt{\left(\sqrt{x+1}+1\right)^2-\left(\sqrt{x+1}+1\right)^2}=\sqrt{0}=0\)
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\(A\left(x\right)=6x^3-x\left(x+2\right)+4\left(x+3\right)\)
\(A\left(x\right)=6x^3-x^2+2x+4x+12\)
\(A\left(x\right)=6x^3-x^2+\left(2x+4x\right)+12\)
\(A\left(x\right)=6x^3-x^2+6x+12\)
\(B\left(x\right)=-x\left(x+1\right)-\left(4-3x\right)+x^2\left(x-2\right)\)
\(B\left(x\right)=-\left(x^2\right)+2-4+3x+x^3-2x^2\)
\(B\left(x\right)=\left(-x^2-2x^2\right)+\left(2-4\right)+3x+x^3\)
\(B\left(x\right)=-3x^2-2+3x+x^3\)
Sửa lại cho Bạn Vũ Đình Phước nhé :v
A (x) = 6x3 – x (x + 2) + 4 (x + 3)
= 6x3 – x2 - 2x + 4x + 12
= 6x3 – x2 + 2x + 12
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a)
1 4 − x = 1 20 + ( − 2 ) 1 4 − x = − 39 20 x = 1 4 − − 39 20 = 5 20 + 39 20 = 44 20 = 11 5
b)
− 2 5 − x = 4 15 + − 1 3 − 2 5 − x = − 1 15 x = − 2 5 − − 1 15 = − 6 15 − − 1 15 = − 5 15 = − 1 3
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a, x= 155 -51 <=> x=104
b.x=-120 : -24 <=> x=5
c.2(81-x)= 155 -35
<=> 2(81-x)=120
<=> 81 - x = 60
<=> x= 21
Lời giải:
a.
$155-x=51$
$x=155-51=104$
b.
$(-24)x=-120$
$x=(-120):(-24)=5$
c.
$35+2(81-x)=155$
$2(81-x)=155-35=120$
$81-x=120:2=60$
$x=81-60=21$
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\(\frac{x}{3}=\frac{y}{4}\rightarrow x=3k;y=4k\)
Thay x=3k; y=4k vào bt có:
\(2\left(3k\right)^2-3\left(4k\right)^2=-120\)
\(\rightarrow18k^2-48k^2=-120\)
\(\rightarrow-30k^2=-120\)
\(\rightarrow k^2=4\)
\(\Rightarrow k=\left\{2;-2\right\}\)
\(\Rightarrow\left[{}\begin{matrix}x=2.3=6\\y=2.4=8\end{matrix}\right.\) hoặc \(\left[{}\begin{matrix}x=-2.3=-6\\y=-2.4=-8\end{matrix}\right.\)