tim xy
a) (2x+1).(y-3)=10
b) x-3=y. (x+2)
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a: \(4x^3-36x\)
\(=4x\cdot x^2-4x\cdot9\)
\(=4x\left(x^2-9\right)=4x\left(x-3\right)\left(x+3\right)\)
b:Sửa đề: \(4x^3-y^3+4x^2y-xy^2\)
\(=4x^2\left(x+y\right)-y^2\left(x+y\right)\)
\(=\left(x+y\right)\left(4x^2-y^2\right)=\left(x+y\right)\left(2x-y\right)\left(2x+y\right)\)
c: \(a^2+2ab-5a-10b\)
=a(a+2b)-5(a+2b)
=(a+2b)(a-5)
d: \(\left(x+1\right)^3-27\)
\(=\left(x+1\right)^3-3^3\)
\(=\left(x+1-3\right)\left\lbrack\left(x+1\right)^2+3\left(x+1\right)+3^2\right\rbrack\)
\(=\left(x-2\right)\left(x^2+2x+1+3x+3+9\right)\)
\(=\left(x-2\right)\left(x^2-5x+13\right)\)
e: \(4xy^2-4x^2y-y^3\)
\(=y\cdot4xy-y\cdot4x^2-y\cdot y^2\)
\(=-y\left(4x^2-4xy+y^2\right)=-y\left(2x-y\right)^2\)
f: \(\left(5x-y\right)^2-\left(x-2y\right)^2\)
=(5x-y-x+2y)(5x-y+x-2y)
=(4x+y)(6x-3y)
=3(2x-y)(4x+y)
g: \(x^3+2x^2+x-16xy^2\)
\(=x\left(x^2+2x+1-16y^2\right)\)
\(=x\left\lbrack\left(x+1\right)^2-\left(4y\right)^2\right\rbrack\)
=x(x+1-4y)(x+1+4y)
a: -2x(x+3)+x(2x-1)=10
=>-2x^2-6x+2x^2-x=10
=>-7x=10
=>x=-10/7
b: Sửa đề: 2/3x(9/2x+1/4)-(3x^2+2)=3
=>3x^2+1/6x-3x^2-2=3
=>1/6x-2=3
=>x=30
\(a,\Leftrightarrow4x^2-24x+36-4x^2+1=10\\ \Leftrightarrow-24x=-27\Leftrightarrow x=\dfrac{9}{8}\\ b,\Leftrightarrow x\left(x^2-25\right)=0\\ \Leftrightarrow x\left(x-5\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
\(a,4.\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)
\(\Leftrightarrow4.\left(x^2-6x+9\right)-\left(2x^2\right)-1^2=10\)
\(\Leftrightarrow4x^2-24x+36-4x^2+1=10\)
\(\Leftrightarrow-24x+27=10\)
\(\Leftrightarrow-24x=-27\)
\(\Leftrightarrow x=\dfrac{27}{24}\)
Vậy \(x=\dfrac{27}{24}\)
2(x-1)-3(2x+2)-4(2x+3)=16
2x-2-6x+6-8x+12=16
(2x-6x-8x)-(2+6+12)=16
(-12).x-20=16
(-12).x=16+20=36
x=36:(-12)=-3
2(x-1)-3(2x+2)-4(2x+3)=16
2x-2-6x-6-8x-12=16
2x-6x-8x=16+2+6+12
-12x=36
x=-3
a, \(3x=5y=7z=>\dfrac{3x}{105}=\dfrac{5y}{105}=\dfrac{7z}{105}=>\dfrac{x}{35}=\dfrac{y}{21}=\dfrac{z}{15}\)
áp dụng tính chất dãy tỉ số = nhau
\(=>\dfrac{x}{35}=\dfrac{y}{21}=\dfrac{z}{15}=\dfrac{x+y+z}{35+21+15}=\dfrac{10}{71}\)
\(=>\dfrac{x}{35}=\dfrac{10}{71}=>x=\dfrac{350}{71}\)
\(=>\dfrac{y}{21}=\dfrac{10}{71}=>y=\dfrac{210}{71}\)
\(=>\dfrac{z}{15}=\dfrac{10}{71}=>z=\dfrac{150}{71}\)
b, \(\)\(6x=5y=>\dfrac{x}{5}=\dfrac{y}{6}=>\dfrac{x}{20}=\dfrac{y}{24}\)
có \(7y=8z=>\dfrac{y}{8}=\dfrac{z}{7}=>\dfrac{y}{24}=\dfrac{z}{21}\)
\(=>\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}=>\dfrac{3x}{60}=\dfrac{2y}{48}=\dfrac{4z}{84}\)
áp dụng t/c dãy tỉ số = nhau
\(=>\dfrac{3x}{60}=\dfrac{2y}{48}=\dfrac{4z}{84}=\dfrac{3x+2y+4z}{60+48+84}=\dfrac{12}{192}=\dfrac{1}{16}\)
\(=>\dfrac{3x}{60}=\dfrac{1}{16}=>x=1,25\)
\(=>\dfrac{2y}{48}=\dfrac{1}{16}=>y=1,5\)
\(=>\dfrac{4z}{84}=\dfrac{1}{16}=>z=1,3125\)
c, \(x:y:z=1:2:3=>\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3}\)
\(=>x=\dfrac{y}{2},z=\dfrac{3y}{2}\)
thay x,z vào \(x^3+y^3+z^3=36=>\left(\dfrac{y}{2}\right)^3+y^3+\left(\dfrac{3y}{2}\right)^3=36\)
\(=>y=2\)
\(=>x=\dfrac{y}{2}=\dfrac{2}{2}=1,z=\dfrac{3y}{2}=\dfrac{3.2}{2}=3\)
d, \(\dfrac{x}{2}=\dfrac{y}{3}=>x=\dfrac{2y}{3}\)
thay x vào \(3x^3+y^3=51=>3.\left(\dfrac{2y}{3}\right)^3+y^3=51=>y=3\)
\(=>x=\dfrac{2.3}{3}=2\)
c, từ đoạn này á
\(\left(\dfrac{y}{2}\right)^3+y^3+\left(\dfrac{3y}{2}\right)^3=36\)
\(< =>\dfrac{y^3}{8}+\dfrac{8y^3}{8}+\dfrac{27y^3}{8}=36\)
\(=>\dfrac{36y^3}{8}=36=>36y^3=8.36=>y^3=8=>y=2\)
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