\(\dfrac{x+1}{1}+\dfrac{2x+3}{3}+\dfrac{3x+5}{5}+...+\dfrac{10x+19}{19}=12+\dfrac{4}{3}+\dfrac{6}{5}+...+\dfrac{20}{19}\)

\(x+1+\dfrac{2x}{3}+1+\dfrac{3x}{5}+1+...+\dfrac{10x}{19}+1-12-\dfrac{4}{3}-\dfrac{6}{5}-...-\dfrac{20}{19}=0\)

\(x+\dfrac{2x}{3}-\dfrac{4}{3}+\dfrac{3x}{5}-\dfrac{6}{5}+...+\dfrac{10x}{19}-\dfrac{20}{19}+10-12=0\)

\(x-2+\dfrac{2x-4}{3}+\dfrac{3x-6}{5}+...+\dfrac{10x-20}{19}=0\)

\(x-2+\dfrac{2\left(x-2\right)}{3}+\dfrac{3\left(x-2\right)}{5}+...+\dfrac{10\left(x-2\right)}{19}=0\)

\(\left(x-2\right)\left(\dfrac{2}{3}+\dfrac{3}{5}+...+\dfrac{10}{19}\right)=0\)

Ta thấy \(\left(\dfrac{2}{3}+\dfrac{3}{5}+...+\dfrac{10}{19}\right)>0\)

\(\Rightarrow x-2=0\Rightarrow x=2\)

a) \(\dfrac{x^4+x^3+x+1}{x^4-x^3+2x^2-x+1}\)

= \(\dfrac{x^3\left(x+1\right)+\left(x+1\right)}{x^3\left(x-1\right)-\left(x-1\right)+2x^2}\)

= \(\dfrac{\left(x+1\right)\left(x^3+1\right)}{\left(x-1\right)\left(x^3-1\right)+2x^2}\)

= \(\dfrac{\left(x+1\right)\left(x+1\right)\left(x^2-x+1\right)}{\left(x-1\right)\left(x-1\right)\left(x^2+x+1\right)+2x^2}\)

= \(\dfrac{\left(x+1\right)^2.\left(x^2-x+1\right)}{\left(x-1\right)^2\left(x^2+x+1\right)+2x^2}\)

Ta thấy mẫu thức của phân thức vốn đã lớn hơn 0 với mọi x, vậy để p/t trên có giá trị bằng 0 thì tử thức phải bằng 0

\(\Rightarrow\left(x+1\right)^2\left(x^2-x+1\right)=0\)

\(\Rightarrow x=-1\)

Vậy x = -1

b) \(\dfrac{x^4-5x^2+4}{x^4-10x^2+9}\)

= \(\dfrac{x^4-x^3+x^3-x^2-4x^2+4}{x^4-x^3+x^3-x^2-9x^2+9}\)

= \(\dfrac{x^3\left(x-1\right)+x^2\left(x-1\right)-4\left(x-1\right)\left(x+1\right)}{x^3\left(x-1\right)+x^2\left(x-1\right)-9\left(x-1\right)\left(x+1\right)}\)

= \(\dfrac{\left(x-1\right)\left(x^3+x^2-4x-4\right)}{\left(x-1\right)\left(x^3+x^2-9x-9\right)}\)

= \(\dfrac{x^3+x^2-4x-4}{x^3+x^2-9x-9}\)

= \(\dfrac{x^2\left(x+1\right)-4\left(x+1\right)}{x^2\left(x+1\right)-9\left(x+1\right)}\)

= \(\dfrac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+1\right)\left(x-3\right)\left(x+3\right)}\)

= \(\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-3\right)\left(x+3\right)}\) ( ĐKXĐ : \(x\ne\pm3\) )

Để phân thức trên có giá trị bằng 0 thì tử thức phải bằng 0

\(\Rightarrow\left(x-2\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\) ( thoả mãn điều kiện xác định )

Vậy x = 2 hoặc x = -2

B1:

a) \(x^3-2x^2+x-2\)

= \(x^2\left(x-2\right)+\left(x-2\right)=\left(x-2\right)\left(x^2+1\right)\)

b) \(2x^3+3x^2-3x-2\)

= \(2x^3-2x^2+5x^2-5x+2x-2\)

= \(2x^2\left(x-1\right)+5x\left(x-1\right)+2\left(x-1\right)\)

= \(\left(x-1\right)\left(2x^2+5x+2\right)\)

= \(\left(x-1\right)\left(2x^2+4x+x+2\right)\)

= \(\left(x-1\right)\left[2x\left(x+2\right)+\left(x+2\right)\right]\)

= \(\left(x-1\right)\left(x+2\right)\left(2x+1\right)\)

c) \(5x^2+5y^2-x^2z+2xyz-y^2z-10xy\)

= \(5\left(x^2+2xy+y^2\right)+z\left(x^2+2xy+y^2\right)\)

= \(5\left(x+y\right)^2+z\left(x+y\right)^2\)

= \(\left(x+y\right)^2\left(5+z\right)\)

d) \(x^3-3x^2y+3xy^2-x+y-y^3\)

= \(\left(x-y\right)^3-\left(x-y\right)\)

= \(\left(x-y\right)\left[\left(x-y\right)^2-1\right]\)

= \(\left(x-y\right)\left(x-y-1\right)\left(x-y+1\right)\)

B2:

a) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)

\(\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)

\(\left(2x-5\right)\left(2x+5-2x-7\right)=0\)

\(\left(2x-5\right).\left(-2\right)=0\)

\(\Rightarrow2x-5=0\Rightarrow x=\dfrac{5}{2}\)

b) \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

\(\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

\(\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)

\(\left(x+3\right)\left(x^2-2x\right)=0\)

\(\left(x+3\right).x.\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=0\\x=2\end{matrix}\right.\)

c) \(2x^3+3x^2+2x+3=0\)

\(x^2\left(2x+3\right)+\left(2x+3\right)=0\)

\(\left(2x+3\right)\left(x^2+1\right)=0\)

Ta thấy \(x^2+1>0\) với mọi x

\(\Rightarrow2x+3=0\Rightarrow x=\dfrac{-3}{2}\)

1)

a) \(x^2-4x+y^2+2y+5=0\)

\(\left(x^2-4x+4\right)+\left(y^2+2x+1\right)=0\)

\(\left(x-2\right)^2+\left(y+1\right)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)

b) \(x^2+2y^2+2xy-2y+1=0\)

\(\left(x^2+2xy+y^2\right)+\left(y^2-2y+1\right)=0\)

\(\left(x+y\right)^2+\left(y-1\right)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}x+y=0\\y-1=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-y\\y=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)

c) \(x^2+2y^2+2xy=2y-2\)

\(x^2+2y^2+2xy-2y+2=0\)

\(\left(x^2+2xy+y^2\right)+\left(y^2-2y+1\right)+1=0\)

\(\left(x+y\right)^2+\left(y-1\right)^2+1=0\)

Ta thấy \(\left\{{}\begin{matrix}\left(x+y\right)^2\ge0\\\left(y-1\right)^2\ge0\\1>0\end{matrix}\right.\)

\(\Rightarrow\) \(\left(x+y\right)^2+\left(y-1\right)^2+1>0\)

Vậy ko có x và y thoả mãn bài toán

\(\left(\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\right).\sqrt{2}\)

= \(\sqrt{2}.\sqrt{2-\sqrt{3}}+\sqrt{2}.\sqrt{2+\sqrt{3}}\)

= \(\sqrt{2.\left(2-\sqrt{3}\right)}+\sqrt{2.\left(2+\sqrt{3}\right)}\)

= \(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\)

= \(\sqrt{3-2\sqrt{3}+1}+\sqrt{3+2\sqrt{3}+1}\)

= \(\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)

= \(\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}\)

nhân số đó lần lượt với 0;1;2;3;..........

a) \(\left(x-y\right)^2=x^2-2xy+y^2=x^2+2xy+y^2-4xy\)

= \(\left(x+y\right)^2-4xy=\left(-7\right)^2-4.12=49-48=1\)

\(\left(x-y\right)^2=1\Rightarrow x-y=1\)

1) Giải

xy + 2 = 2x + y

xy + 2 - 2x - y = 0

x ( y - 2 ) - ( y - 2 ) = 0

( y - 2 ).( x - 1 ) = 0

\(\Rightarrow\left\{{}\begin{matrix}y=2\\x=1\end{matrix}\right.\)

2) Giải:

Ta có: \(a^2+b^2=c^2+d^2\)

\(\Rightarrow\) \(a^2-c^2=d^2-b^2\)

\(\left(a-c\right)\left(a+c\right)=\left(d-b\right)\left(d+b\right)\) (*)

Ta có: \(a+b=c+d\) (**)

\(\Rightarrow a-c=b-d\)

+) Nếu \(a-c=0\)

\(\Rightarrow a=c\) và \(b=d\)

Nên \(a^{2010}+b^{2010}=c^{2010}+d^{2010}\)

+) Nếu \(a-c\ne0\) và \(b-d\ne0\)

thì \(a\ne c\) và \(b\ne d\)

Khi đó (*) \(\Leftrightarrow\) \(a+c=b+d\) (***)

Cộng (**) và (***) theo vế:

2a + b + c = 2d + b + c

2a = 2d

a = d

Suy ra b = c

Do đó \(a^{2010}+b^{2010}=c^{2010}+d^{2010}\)

1) tìm GTNN

a) \(B=\left|x-2017\right|+\left|x-20\right|\)

B \(\ge\left|x-2017-x+20\right|=\left|-1997\right|=1997\)

Dấu " = " xảy ra khi và chỉ khi 20 \(\le x\le2017\)

Vậy Min_B = 1997 khi 20 \(\le x\le2017\)

b) \(C=\left|x-3\right|+\left|x-5\right|\)

\(C\ge\left|x-3-x+5\right|=\left|2\right|=2\)

Dấu " = " xảy ra khi 3 \(\le x\le5\)

Vậ Min_C = 2 khi và chỉ khi 3 \(\le x\le5\)

c) \(C=\left|x^2+4\right|+3\)

Ta thấy \(x^2+4\ge0\) với mọi x

nên \(\left|x^2+4\right|+3=x^2+4+3=x^2+7\)\(\ge\) 7

Dấu " =" xảy ra khi x = 0

Min_C = 7 khi và chỉ khi x = 0

Tóm tắt: R₁ ; R₂

UĐ = 10V

I₁ = 0,91A

I₂ = 0,36A

U = 220V => có thể mắc R₁ và R₂ nối tiếp không?

Giải:

Điện trở của mỗi đèn lần lượt là:

R₁ = \(\dfrac{U_Đ}{I_1}=\dfrac{10}{0,91}\approx11\) \(\Omega\)

R₂ = \(\dfrac{U_Đ}{I_2}=\dfrac{10}{0,36}\approx28\Omega\)

Cường độ dòng điện qua hai đèn khi mắc R₁ nt R₂ là:

I = \(\dfrac{U}{R_{tđ}}=\dfrac{U}{R_1+R_2}=\dfrac{220}{11+28}\approx5,64\)A

Ta thấy I > I₁ nên Đ₁ sáng quá mức \(\Rightarrow\) đứt dây tóc

Tương tự, I > I₂ nên Đ₂ sáng quá mức \(\Rightarrow\) đứt dây tóc