20,23 : 0,01 + 2,023 x 4 x 20% x 250 + 202,3 : 10% - 2023 x 1/5
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\(x\times\dfrac{1}{4}+x\times75\%=20,23\times90+202,3\)
\(x\times\dfrac{1}{4}+x\times\dfrac{3}{4}=20,23\times90+20,23\times10\)
\(x\times\left(\dfrac{1}{4}+\dfrac{3}{4}\right)=20,23\times\left(90+10\right)\)
\(x\times1=20,23\times100\)
\(x=2023\)
X x 1/4+ X x 75% = 20,23 x 90 +202,3
X x 1/4+ X x 3/4 = 2023
X x ( 1/4 +3/4 ) = 2023
X x 1 = 2023
x = 2023
\(\text{25 x 20,23 + 105 x 20,23 - 20,25 x 20,23 - 9,75 x 20,23}\\ =20,23\text{ x }\left(25+105-9,75-20,25\right)\\ =20.23\text{ x }100=2023\)
\(\dfrac{5}{13}\times\dfrac{7}{11}+\dfrac{5}{13}\times\dfrac{2}{11}+\dfrac{8}{13}\times\dfrac{9}{11}\)
\(=\dfrac{5}{13}\times\left(\dfrac{7}{11}+\dfrac{2}{11}\right)+\dfrac{8}{13}\times\dfrac{9}{11}\)
\(=\dfrac{5}{13}\times\dfrac{9}{11}+\dfrac{8}{13}\times\dfrac{9}{11}\)
\(=\dfrac{9}{11}\times\left(\dfrac{5}{13}+\dfrac{8}{13}\right)\)
\(=\dfrac{9}{11}\times1\)
\(=\dfrac{9}{11}\)
a) \(20,23\times21+20,23\times80-20,23\)
\(=20,23\times\left(21+80-1\right)\)
\(=20,23\times100\)
\(=2023\)
`#3107.101107`
`202,3 \times 7,5 + 202,3 + 202,3 + 202,3 \div 2`
`= 202,3 \times 7,5 + 202,3 \times 2 + 202,3 \times` $\dfrac{1}2$
`= 202,3 \times (7,5 + 2 +`$\dfrac{1}2)$
`= 202,3 \times (7,5 + 2 + 0,5)`
`= 202,3 \times 10`
`= 2023`
\(2,023\times570+2,023\times575-145\times2,023\)
\(=2,023\times\left(570+575-145\right)\)
\(=2,023\times1000\)
\(=2023\)
Bài 1
a) 5x²y - 20xy²
= 5xy(x - 4y)
b) 1 - 8x + 16x² - y²
= (1 - 8x + 16x²) - y²
= (1 - 4x)² - y²
= (1 - 4x - y)(1 - 4x + y)
c) 4x - 4 - x²
= -(x² - 4x + 4)
= -(x - 2)²
d) x³ - 2x² + x - xy²
= x(x² - 2x + 1 - y²)
= x[(x² - 2x+ 1) - y²]
= x[(x - 1)² - y²]
= x(x - 1 - y)(x - 1 + y)
= x(x - y - 1)(x + y - 1)
e) 27 - 3x²
= 3(9 - x²)
= 3(3 - x)(3 + x)
f) 2x² + 4x + 2 - 2y²
= 2(x² + 2x + 1 - y²)
= 2[(x² + 2x + 1) - y²]
= 2[(x + 1)² - y²]
= 2(x + 1 - y)(x + 1 + y)
= 2(x - y + 1)(x + y + 1)
Bài 2:
a: \(x^2\left(x-2023\right)+x-2023=0\)
=>\(\left(x-2023\right)\left(x^2+1\right)=0\)
mà \(x^2+1>=1>0\forall x\)
nên x-2023=0
=>x=2023
b:
ĐKXĐ: x<>0
\(-x\left(x-4\right)+\left(2x^3-4x^2-9x\right):x=0\)
=>\(-x\left(x-4\right)+2x^2-4x-9=0\)
=>\(-x^2+4x+2x^2-4x-9=0\)
=>\(x^2-9=0\)
=>(x-3)(x+3)=0
=>\(\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
c: \(x^2+2x-3x-6=0\)
=>\(\left(x^2+2x\right)-\left(3x+6\right)=0\)
=>\(x\left(x+2\right)-3\left(x+2\right)=0\)
=>(x+2)(x-3)=0
=>\(\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
d: 3x(x-10)-2x+20=0
=>\(3x\left(x-10\right)-\left(2x-20\right)=0\)
=>\(3x\left(x-10\right)-2\left(x-10\right)=0\)
=>\(\left(x-10\right)\left(3x-2\right)=0\)
=>\(\left[{}\begin{matrix}x-10=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=10\end{matrix}\right.\)
Câu 1:
a: \(5x^2y-20xy^2\)
\(=5xy\cdot x-5xy\cdot4y\)
\(=5xy\left(x-4y\right)\)
b: \(1-8x+16x^2-y^2\)
\(=\left(16x^2-8x+1\right)-y^2\)
\(=\left(4x-1\right)^2-y^2\)
\(=\left(4x-1-y\right)\left(4x-1+y\right)\)
c: \(4x-4-x^2\)
\(=-\left(x^2-4x+4\right)\)
\(=-\left(x-2\right)^2\)
d: \(x^3-2x^2+x-xy^2\)
\(=x\left(x^2-2x+1-y^2\right)\)
\(=x\left[\left(x^2-2x+1\right)-y^2\right]\)
\(=x\left[\left(x-1\right)^2-y^2\right]\)
\(=x\left(x-1-y\right)\left(x-1+y\right)\)
e: \(27-3x^2\)
\(=3\left(9-x^2\right)\)
\(=3\left(3-x\right)\left(3+x\right)\)
f: \(2x^2+4x+2-2y^2\)
\(=2\left(x^2+2x+1-y^2\right)\)
\(=2\left[\left(x^2+2x+1\right)-y^2\right]\)
\(=2\left[\left(x+1\right)^2-y^2\right]\)
\(=2\left(x+1+y\right)\left(x+1-y\right)\)
Lời giải:
$2,5\times 20,23\times 8\times 5=2,5\times 20,23\times 4\times 2\times 5$
$=(2,5\times 4)\times 20,23\times (2\times 5)$
$=10\times 20,23\times 10=20,23\times 100=2023$
202,3 x0,86 + 0,14 x606,9 + 0,85 x 404,6 +202,3 x7,02
= 202,3 x 0,86 + 0,14 x 3 x 202,3 + 0,85 x 2 x 202,3 + 202,3 x 7,02
= 202,3 x 0,86 + 0,42 x 202,3 + 1,7 x 202,3 + 202,3 x 7,02
= 202,3 x (0,86 + 0,42 + 1,7+ 7,02)
= 202,3 x 10= 2023
20,23:0,01+2,023x4x20%x250+202,3:10%-2023x1/5
\(=20,23\times100+2,023\times200+202,3\times10-2023\times0,2\)
\(=2023+202,3\times2+2023-2023\times0,2\)
=2023+2023
=4046