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\(2x+3y+5z=\frac{x^2+y^2+z^2}{2}+19\)
\(x^2+y^2+z^2+38=4x+6y+10z\)
\(\left(x^2-4x+4\right)+\left(y^2-6y+9\right)+\left(z^2-10z+25\right)=0\)
\(\left(x-2\right)^2+\left(y-3\right)^2+\left(z-5\right)^2=0\)
\(x-2=y-3=z-5=0\)
\(x=2,y=3,z=5\)
Bài 1:
Vận tốc cano khi dòng nước lặng là: $25-2=23$ (km/h)
Bài 2:
Đổi 1 giờ 48 phút = 1,8 giờ
Độ dài quãng đường AB: $1,8\times 25=45$ (km)
Vận tốc ngược dòng là: $25-2,5-2,5=20$ (km/h)
Cano ngược dòng từ B về A hết:
$45:20=2,25$ giờ = 2 giờ 15 phút.
Bài 1:
a.
$a^3-a^2c+a^2b-abc=a^2(a-c)+ab(a-c)$
$=(a-c)(a^2+ab)=(a-c)a(a+b)=a(a-c)(a+b)$
b.
$(x^2+1)^2-4x^2=(x^2+1)^2-(2x)^2=(x^2+1-2x)(x^2+1+2x)$
$=(x-1)^2(x+1)^2$
c.
$x^2-10x-9y^2+25=(x^2-10x+25)-9y^2$
$=(x-5)^2-(3y)^2=(x-5-3y)(x-5+3y)$
d.
$4x^2-36x+56=4(x^2-9x+14)=4(x^2-2x-7x+14)$
$=4[x(x-2)-7(x-2)]=4(x-2)(x-7)$
Bài 2:
a. $(3x+4)^2-(3x-1)(3x+1)=49$
$\Leftrightarrow (3x+4)^2-[(3x)^2-1]=49$
$\Leftrightarrow (3x+4)^2-(3x)^2=48$
$\Leftrightarrow (3x+4-3x)(3x+4+3x)=48$
$\Leftrightarrow 4(6x+4)=48$
$\Leftrightarrow 6x+4=12$
$\Leftrightarrow 6x=8$
$\Leftrightarrow x=\frac{4}{3}$
b. $x^2-4x+4=9(x-2)$
$\Leftrightarrow (x-2)^2=9(x-2)$
$\Leftrightarrow (x-2)(x-2-9)=0$
$\Leftrightarrow (x-2)(x-11)=0$
$\Leftrightarrow x-2=0$ hoặc $x-11=0$
$\Leftrightarrow x=2$ hoặc $x=11$
c.
$x^2-25=3x-15$
$\Leftrightarrow (x-5)(x+5)=3(x-5)$
$\Leftrightarrow (x-5)(x+5-3)=0$
$\Leftrightarrow (x-5)(x+2)=0$
$\Leftrightarrow x-5=0$ hoặc $x+2=0$
$\Leftrightarrow x=5$ hoặc $x=-2$
x^2 - x - y^2 - y
= x^2 - y^2 - x - y
= ( x - y ) ( x + y ) - ( x + y )
= ( x + y ) ( x - y - 1 )
x^2 - 2xy + y^2 - z^2
= ( x- y ) ^2 - z^2
= ( x - y - z ) ( x - y + z )
\(a,A=\dfrac{2^2-9}{3\left(2+5\right)}=\dfrac{-5}{21}\\ b,B=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x-9}{\left(x-3\right)\left(x+3\right)}\\ B=\dfrac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\\ c,P=AB=\dfrac{\left(x-3\right)\left(x+3\right)}{3\left(x+5\right)}\cdot\dfrac{3}{x+3}=\dfrac{x-3}{x+5}\\ P=\dfrac{x+5-8}{x+5}=1-\dfrac{8}{x+5}\in Z\\ \Leftrightarrow x+5\inƯ\left(8\right)=\left\{-8;-4;-2;-1;1;2;4;8\right\}\\ \Leftrightarrow x\in\left\{-13;-9;-7;-6;-4;-1\right\}\left(x\ne\pm3\right)\)
\(a.\) \(Thay\) \(x=2\left(TM\right):\) \(\dfrac{2^2-9}{3\left(2+5\right)}=\dfrac{-5}{21}.\)
\(b.\) \(B=\dfrac{x}{x+3}+\dfrac{2x}{x-3}-\dfrac{3x^2+9}{x^2-9}.\) \(\left(x\ne-5;x\ne3;x\ne-3\right).\)
\(B=\dfrac{x\left(x-3\right)+2x\left(x+3\right)-3x^2-9}{\left(x-3\right)\left(x+3\right)}.\)
\(B=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x-9}{\left(x-3\right)\left(x+3\right)}.\)
\(B=\dfrac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}.\)
\(c.\) \(P=A.B.\Rightarrow P=\dfrac{x^2-9}{3\left(x+5\right)}.\dfrac{3}{x+3}=\dfrac{x-3}{x+5}=1+\dfrac{-8}{x+5}.\)
Để \(P\in Z.\Leftrightarrow1+\dfrac{-8}{x+5}\in Z.\Leftrightarrow x+5\in\) Ư \(\left(-8\right)=\left(1;-1;2;-2;4;-4;8;-8\right).\)
\(\Rightarrow x\in\left\{-4;-6;-7;-1;-9;-13\right\}.\)