b,\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)

=>\(\dfrac{bc}{abc}+\dfrac{ac}{bac}+\dfrac{ab}{abc}=0\)

=>\(\dfrac{ab+ac+bc}{abc}=0\)

=>ab+ac+bc=0

=>ab=-ac-bc

ac=-ab-bc

bc=-ab-ac

N=\(\dfrac{1}{a^2+2bc}+\dfrac{1}{b^2+2ca}+\dfrac{1}{c^2+2ab}\)

N=\(\dfrac{1}{a^2+bc+bc}+\dfrac{1}{b^2+ca+ca}+\dfrac{1}{c^2+ab+ab}\)

N=\(\dfrac{1}{a^2-ab-ac+bc}+\dfrac{1}{b^2-ab-bc+ca}+\dfrac{1}{c^2-ac-bc+ab}\)

N=\(\dfrac{1}{a\left(a-b\right)-c\left(a-b\right)}+\dfrac{1}{b\left(b-a\right)-c\left(b-a\right)}+\dfrac{1}{c\left(c-a\right)-b\left(c-a\right)}\)

N=\(\dfrac{1}{\left(a-c\right)\left(a-b\right)}+\dfrac{1}{\left(b-c\right)\left(b-a\right)}+\dfrac{1}{\left(c-b\right)\left(c-a\right)}\)

N=\(\dfrac{b-c}{\left(a-c\right)\left(b-c\right)\left(a-b\right)}-\dfrac{a-c}{\left(b-c\right)\left(a-b\right)\left(a-c\right)}+\dfrac{a-b}{\left(b-c\right)\left(a-c\right)\left(a-b\right)}\)

N=\(\dfrac{b-c-a+c+a-b}{\left(a-c\right)\left(b-c\right)\left(a-b\right)}\)=0

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

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

b,

thay x=\(\dfrac{1}{2}\) vào bt M ta được:

\(1-\left(\dfrac{1}{2}\right)^3-\left(\dfrac{1}{2}\right)^2=\dfrac{5}{8}\)

Bài 2 hình như sai đề thì phải

a) Ta có: \(P=\left(\frac{\sqrt{a}}{\sqrt{a}-1}+\frac{\sqrt{a}}{a-1}\right):\left(\frac{2}{a}-\frac{2-a}{a\sqrt{a}+a}\right)\)

\(=\left(\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right):\left(\frac{2\left(\sqrt{a}+1\right)}{a\left(\sqrt{a}+1\right)}-\frac{2-a}{a\left(\sqrt{a}+1\right)}\right)\)

\(=\frac{a+\sqrt{a}+\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}:\frac{2\sqrt{a}+2-2+a}{a\left(\sqrt{a}+1\right)}\)

\(=\frac{a+2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\cdot\frac{a\left(\sqrt{a}+1\right)}{a+2\sqrt{a}}\)

\(=\frac{a}{\sqrt{a}-1}\)

b)

ĐKXĐ: \(a\notin\left\{1;0\right\}\)

Để P-2 là số dương thì P-2>0

⇔\(\frac{a}{\sqrt{a}-1}-2>0\)

\(\Leftrightarrow\frac{a}{\sqrt{a}-1}-\frac{2\left(\sqrt{a}-1\right)}{\sqrt{a}-1}>0\)

\(\Leftrightarrow\frac{a-2\sqrt{a}+2}{\sqrt{a}-1}>0\)

mà \(a-2\sqrt{a}+2=\left(\sqrt{a}-1\right)^2+1>0\forall a\)

nên \(\sqrt{a}-1>0\)

\(\Leftrightarrow\sqrt{a}>1\)

\(\Leftrightarrow a>1\)(tm)

Vậy: Khi a>1 thì P-2 là số dương

A=\((\frac{\sqrt{a}\left(\sqrt{a}+1\right)+\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}):\left(\frac{2\left(\sqrt{a}+1\right)-\left(2-a\right)}{a\left(\sqrt{a}+1\right)}\right)\)

\(A=\left(\frac{a+\sqrt{a}+\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right):\left(\frac{2\sqrt{a}+2-2+a}{a\left(\sqrt{a}+1\right)}\right)\)

\(A=\frac{a+2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}.\frac{a\left(\sqrt{a}+1\right)}{2\sqrt{a}-a}\)

\(A=\frac{a}{\sqrt{a}-1}\)

\(M=\dfrac{-1}{3}.\left(-\left(x^4\right)\right).\left(y^3\right)\)

Bậc của đơn thức M là : 7

Hệ số của M : \(\dfrac{-1}{3}\)

b) \(M=\dfrac{-1}{3}.\left(-\left(-2^4\right)\right).2^3\)

\(M=\dfrac{-1}{3}.\left(-16\right).8=\dfrac{128}{3}\)

Mink ko biết dúng hay sai nha @Cao Chu Thiên Trang

Sửa lại nha :

\(M=\left(\dfrac{-2}{3}.\dfrac{1}{2}.-1\right)\left(x^2.x.x\right)\left(y.y^2\right)\)

=> \(M=\dfrac{1}{3}.x^4.y^3\)

Bậc của đơn thức là 7

Hệ số là \(\dfrac{1}{3}\)

b) \(M=\dfrac{1}{3}.\left(-2\right)^4.2^3\)\(=\dfrac{128}{3}\)

Bài 2:

a: Để (d) tạo với trục Ox một góc nhọn thì 1-2m>0

=>2m<1

=>m<1/2

b: y=(1-2m)x+m-1

=x-2mx+m-1

=>x-2mx+m-1-y=0

=>m(-2x+1)+x-y-1=0

Điểm mà (d) luôn đi qua có tọa độ là:

-2x+1=0 và x-y=1

=>x=1/2 và y=x-1=1/2-1=-1/2

c: \(d\left(O;d\right)=\dfrac{\left|\left(1-2m\right)\cdot0+\left(-1\right)\cdot0+m-1\right|}{\sqrt{\left(1-2m\right)^2+1}}=\dfrac{\left|m-1\right|}{\sqrt{\left(2m-1\right)^2+1}}\)

Để d lớn nhất thì \(\sqrt{\left(2m-1\right)^2+1}_{MIN}\)

=>m=1/2