Bài 1:

a) \(x^2-y^2+10x+25\)

\(=\left(x^2+10x+25\right)-y^2\)

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

\(=\left(x+y+5\right)\left(x-y+5\right)\)

b) \(x^3-x^2-5x+125\)

\(=x^3+5x^2-6x^2-30x+25x+125\)

\(=x^2\left(x+5\right)-6x\left(x+5\right)+25\left(x+5\right)\)

\(=\left(x+5\right)\left(x^2-6x+25\right)\)

c) \(x^4+4y^4\)

\(=\left(x^2\right)^2+2x^22y^2+\left(2y^2\right)^2-2x^22y^2\)

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

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

d)Sửa đề \(a\left(b^2-c^2\right)+b\left(c^2-a^2\right)+c\left(a^2-b^2\right)\)

\(=a\left(b^2-c^2\right)-b\left[\left(b^2-c^2\right)+\left(a^2-b^2\right)\right]+c\left(a^2-b^2\right)\)

\(=a\left(b^2-c^2\right)-b\left(b^2-c^2\right)-b\left(a^2-b^2\right)+c\left(a^2-b^2\right)\)

\(=\left(a-b\right)\left(b^2-c^2\right)-\left(b-c\right)\left(a^2-b^2\right)\)

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

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

\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

e) \(7x^2-10xy+3y^2\)

\(=\left(\sqrt{7}x\right)^2-2.\sqrt{7}x.\sqrt{3}y+\left(\sqrt{3}y\right)^2\)

\(=\left(\sqrt{7}x-\sqrt{3}y\right)^2\)

f) Sửa đề \(a^3+b^3+c^3-3abc\)

\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc+2ab-3ab\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc-ab\right)\)

h) \(xy\left(x+y\right)-yz\left(y+z\right)+xz\left(x-z\right)\)

\(=x^2y+xy^2-y^2z-yz^2+x^2z-xz^2\)

\(=\left(x^2y+x^2z\right)+\left(xy^2-xz^2\right)-yz\left(y+z\right)\)

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

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

\(=\left(y+z\right)\left[x^2+x\left(y-z\right)-yz\right]\)

\(=\left(y+z\right)\left(x^2+xy-xz-yz\right)\)

\(=\left(y+z\right)\left[x\left(x+y\right)-z\left(x+y\right)\right]\)

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

Vì \(x^{2016}+y^{2016}=x^{2017}+y^{2017}=x^{2018}+y^{2018}\left(x,y\ge0\right)\)

\(\Rightarrow x=y=1\)

\(\Rightarrow A=1^{2019}+1^{2019}\)

\(\Rightarrow A=2\)

Vì \(x^{2015}+y^{2015}=x^{2016}+y^{2016}=x^{2017}+y^{2017}\)

\(\Rightarrow x=y=1\) hoặc \(x=y=0\)

Với \(x=y=1\)

\(S=2018\left(1^{2018}+1^{2018}\right)\)

\(S=2018.2\)

\(S=4036\)

Với \(x=y=0\)

\(S=2018\left(0^{2018}+0^{2018}\right)\)

\(S=0\)

d) \(D=x^2+2xy+y^2-4x-4y+1\)

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

\(D=\left(x+y\right)\left(x+y-4\right)+1\)

Thay x + y = 3 vào D

\(D=3\left(3-4\right)+1\)

\(D=-2\)

c) Ta có:

\(a+b=9\)

\(a-b=3\)

\(\Rightarrow2a=12\)

\(\Rightarrow a=6\)

\(\Rightarrow b=3\)

Thay a = 6 và b = 3 vào C

\(C=a^3-b^3\)

\(C=6^3-3^3\)

\(C=216-27\)

\(C=189\)

a) Ta có:

\(x+y=2\)

\(\Rightarrow\left(x+y\right)^2=4\)

\(\Rightarrow x^2+y^2+2xy=4\)

\(\Rightarrow10+2xy=4\) ( Vì x² + y² = 10 )

\(\Rightarrow2xy=4-10=-6\)

\(\Rightarrow xy=-3\)

\(A=x^3+y^3\)

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

Thay x + y = 2 ; xy = -3 ; x² + y² = 10 vào A

\(A=2\left[10-\left(-3\right)\right]\)

\(A=2.13=26\)

Bài 2:

Ta có:

\(2x=3y=6z\)

\(=\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{6}}\)

\(=\dfrac{x+y+z}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}}\) ( Áp dụng tính chất dãy tỉ số bằng nhau )

\(=\dfrac{1830}{1}=1830\)

Với \(\left\{{}\begin{matrix}2x=1830\\3y=1830\\6z=1830\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=915\\y=610\\z=305\end{matrix}\right.\)

Ta có:

\(A-B=2m^3+3n^3-4mn^2\)

TH1: Nếu m > n. Đặt m = n + x

\(A-B=2\left(n+x\right)^3+3n^3-4\left(n+x\right)n^2\)

\(A-B=2\left(n^3+3n^2x+3nx^2+x^3\right)+3n^3-4n^3-4n^2x\)

\(A-B=2n^3+6n^2x+6nx^2+2x^3+3n^3-4n^3-4n^2x\)

\(A-B=n^3+2n^2x+6nx^2+2x^3>0\)

\(\Rightarrow A>B\)

TH2: Nếu m < n. Đặt n = m + y

Làm tương tự ra được A > B

Bài 1:

a) \(4\dfrac{1}{3}:\dfrac{x}{4}=6:0,3\)

\(\Rightarrow\dfrac{13}{3}.\dfrac{4}{x}=20\)

\(\Rightarrow\dfrac{52}{3x}=20\)

\(\Rightarrow52=20.3x\)

\(\Rightarrow60x=52\)

\(\Rightarrow x=\dfrac{13}{15}\)

b) \(\left(2^3:2^4\right).2^{x+1}=64\)

\(\Rightarrow2^{3-4}.2^{x+1}=64\)

\(\Rightarrow2^{-1}.2^{x+1}=64\)

\(\Rightarrow2^{-1+x+1}=64\)

\(\Rightarrow2^x=64\)

\(\Rightarrow2^x=2^6\)

\(\Rightarrow x=6\)

c) \(\left(x-1\right)^5=-32\)

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

\(\Rightarrow x-1=-2\)

\(\Rightarrow x=-2+1=-1\)

d) \(|3-2x|-3=-3\)

\(\Rightarrow|3-2x|=-3+3=0\)

\(\Rightarrow3-2x=0\)

\(\Rightarrow2x=3\)

\(\Rightarrow x=\dfrac{3}{2}\)

e) \(|x+\dfrac{4}{5}|-\dfrac{1}{7}=0\)

\(\Rightarrow|x+\dfrac{4}{5}|=\dfrac{1}{7}\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{4}{5}=\dfrac{1}{7}\\x+\dfrac{4}{5}=-\dfrac{1}{7}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{7}-\dfrac{4}{5}\\x=-\dfrac{1}{7}-\dfrac{4}{5}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{23}{35}\\x=-\dfrac{33}{35}\end{matrix}\right.\)

Ta có:

\(2\left(a^2+b^2\right)=\left(a+b\right)^2\)

\(\Rightarrow2a^2+2b^2=a^2+2ab+b^2\)

\(\Rightarrow2a^2+2b^2-a^2-2ab-b^2=0\)

\(\Rightarrow a^2-2ab+b^2=0\)

\(\Rightarrow\left(a-b\right)^2=0\)

\(\Rightarrow a-b=0\)

\(\Rightarrow a=b\)