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NT
0
NT
Nguyễn Tuấn Tú
CTVHS
26 tháng 6
\(3x^3-14x^2+4x+3\)
\(=\left(3x^3-15x^2+9x\right)+\left(x^2-5x+3\right)\)
\(=3x\left(x^2-5x+3\right)+\left(x^2-5x+3\right)\)
\(=\left(3x+1\right)\left(x^2-5x+3\right)\)
NT
1
NT
Nguyễn Tuấn Tú
CTVHS
26 tháng 6
Ta có: \(\dfrac{A+C+E}{3}+\dfrac{A+B+D}{3}=40+28\)
\(\Rightarrow\dfrac{2A+B+C+D+E}{3}=68\)
\(\Rightarrow\dfrac{2A}{3}+\dfrac{B+C+D+E}{3}=68\)
Thay \(\dfrac{B+C+D+E}{3}=33\) được:
\(\dfrac{2A}{3}+33=68\)
\(\Rightarrow\dfrac{2}{3}A=68-33\)
\(\Rightarrow\dfrac{2}{3}A=35\)
\(\Rightarrow A=35:\dfrac{2}{3}\)
\(\Rightarrow A=\dfrac{105}{2}=52,5\)
Vậy \(A=52,5\)
26 tháng 6
Bài 22:
A={x\(\in N\)|0<x<=5}
Bài 21: C={x\(\in\)N|7<=x<=14}
Bài 20: B={\(x\in\)N|7<x<17}
Bài 19: A={x\(\in\)N|x>=8}
Bài 18:
C={x\(\in\)N|x>11}
bài 17:
B={\(x\in\)N|x<8}
Bài 16:
A={x\(\in\)N|x<3}
NT
Nguyễn Tuấn Tú
CTVHS
26 tháng 6
\(\dfrac{2^{17}\cdot9^4}{6^3\cdot8^3}\)
\(=\dfrac{2^{17}\cdot3^8}{2^3\cdot3^3\cdot2^9}\)
\(=\dfrac{2^{17}\cdot3^8}{2^{12}\cdot3^3}\)
\(=2^5\cdot3^5\)
\(=6^5=7776\)
NT
Nguyễn Tuấn Tú
CTVHS
26 tháng 6
\(\left(x-1\right)^2+\left(y+1\right)^2=0\)
Nhận xét:
\(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\\\left(y+1\right)^2\ge0\end{matrix}\right.\)
\(\Rightarrow\left(x-1\right)^2+\left(y+1\right)^2\ge0\)
Do đó: Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}x-1=0\\y+1=0\end{matrix}\right.\)
\(\Rightarrow x=1;y=-1\)
Vậy \(x=1;y=-1\)
NT
2
H9
HT.Phong (9A5)
CTVHS
26 tháng 6
\(\left(5x+1\right)^2-\left(2xy-3\right)^2\\ =\left[\left(5x+1\right)-\left(2xy-3\right)\right]\left[\left(5x+1\right)+\left(2xy-3\right)\right]\\ =\left(5x+1-2xy+3\right)\left(5x+1+2xy-3\right)\\ =\left(5x-2xy+4\right)\left(5x+2xy-2\right)\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
26 tháng 6
(5\(x\) + 1)2 - (2\(xy\) - 3)2
= [(5\(x\) + 1) - (2\(xy\) - 3)].[(5\(x\) + 1) + (2\(xy\) - 3)]
= [ 5\(x\) + 1 - 2\(xy\) + 3][5\(x\) + 1 + 2\(xy\) - 3]
= [5\(x\) - 2\(xy\) + (1 + 3)][5\(x\) + 2\(xy\) - (3 - 1)]
= [5\(x\) - 2\(x\)\(y\) + 4][5\(x+2xy\) - 2]
Bài 1 :
\(i,25.\left(x-4\right)=0\)
\(x-4=0:25\)
\(x-4=0\)
\(x=0+4\)
\(x=4\)
Vậy .....
\(m,34.\left(2x-6\right)=0\)
\(2x-6=0:34\)
\(2x-6=0\)
\(2x=0+6\)
\(2x=6\)
\(x=6:2\)
\(x=3\)
Vậy ....
\(n,2023.\left(3x-12\right)=0\)
\(3x-12=0:2023\)
\(3x-12=0\)
\(3x=0+12\)
\(3x=12\)
\(x=12:3\)
\(x=4\)
Vậy...
\(o,47.\left(5x-15\right)=0\)
\(5x-15=0:47\)
\(5x-15=0\)
\(5x=0+15\)
\(5x=15\)
\(x=15:5\)
\(x=3\)
Vậy....
\(p,13.\left(4x-24\right)=0\)
\(4x-24=0:13\)
\(4x-24=0\)
\(4x=0+24\)
\(4x=24\)
\(x=24:4\)
\(x=6\)
Vậy...
\(s,2.\left(x-5\right)-17=25\)
\(2.\left(x-5\right)=25+17\)
\(2.\left(x-5\right)=42\)
\(x-5=42:2\)
\(x-5=21\)
\(x=21+5\)
\(x=26\)
Vậy...
\(t,3.\left(x+7\right)-15=27\)
\(3.\left(x+7\right)=27+15\)
\(3.\left(x+7\right)=42\)
\(x+7=42:3\)
\(x+7=14\)
\(x=14-7\)
\(x=7\)
Vậy...
\(u,15+4.\left(x-2\right)=95\)
\(4.\left(x-2\right)=95-15\)
\(4.\left(x-2\right)=80\)
\(x-2=80:4\)
\(x-2=20\)
\(x=20+2\)
\(x=22\)
Vậy...
\(w,24+3.\left(5-x\right)=27\)
\(3.\left(5-x\right)=27-24\)
\(3.\left(x-5\right)=3\)
\(x-5=3:3\)
\(x-5=1\)
\(x=1+5\)
\(x=6\)
Vậy...
Bài 2 :
\(a,\left(x-2021\right).958=0\)
\(x-2021=0:958\)
\(x-2021=0\)
\(x=0+2021\)
\(x=2021\)
Vậy...
\(b,959.\left(x-7\right)=0\)
\(x-7=0:959\)
\(x-7=0\)
\(x=0+7\)
\(x=7\)
Vậy....
\(e,45.\left(91-x\right)=90\)
\(91-x=90:45\)
\(91-x=2\)
\(x=91-2\)
\(x=89\)
Vậy...
\(g,5x+73.21=73.26\)
\(5x+1533=1898\)
\(5x=1898-1533\)
\(5x=365\)
\(x=365:5\)
\(x=73\)
Vậy...
\(h,\left(x-12\right).105=525\)
\(x-12=525:105\)
\(x-12=5\)
\(x=5+12\)
\(x=17\)
Vậy...
\(i,47.\left(27-x\right)=47\)
\(27-x=47:47\)
\(27-x=1\)
\(x=27-1\)
\(x=26\)
Vậy ...
\(j,2x+69.2=69.4\)
\(2x+138=276\)
\(2x=276-138\)
\(2x=138\)
\(x=138:2\)
\(x=69\)
Vậy ...
\(l,\left(x-40\right).15=15.3\)
\(\left(x-40\right).15=45\)
\(x-40=45:15\)
\(x-40=3\)
\(x=3+40\)
\(x=43\)
Vậy...