T^T help me

I don't now

...............

.................

.

(2x - 1)^2 + (x + 3)^2 - 5(x + 7)(x - 7) = 0
<=>4x^2-4x+1+x^2+6x+9-5x^2+245=0
<=>2x+255=0
<=>2x=-255
<=>x=-255/2

Có trên google ( ghi nguồn đầy đủ )

\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(\Leftrightarrow\)\(4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)

\(\Leftrightarrow\)\(5x^2+2x+10-5x^2+245=0\)

\(\Leftrightarrow\)\(2x=-255\)

\(\Leftrightarrow\)\(x=-127,5\)

Vậy...

Bài 2:

\(M=x^2-2xy+y^2=\left(x-y\right)^2=\left(-3\right)^2=9\)

\(N=x^2+y^2=\left(x-y\right)^2+2xy=9+2.10=29\)

\(P=x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3=\left(-3\right)^3=-27\)

\(Q=x^3-y^3=\left(x-y\right)^3+3xy\left(x-y\right)=\left(-3\right)^3+3.10.\left(-3\right)=-117\)

Bài 1:

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

b)  \(B=x^2+y^2=\left(x+y\right)^2-2xy=\left(-1\right)^2-2.\left(-12\right)=25\)

c)  \(C=x^3+3x^2y+3xy^2+y^3=\left(x+y\right)^3=\left(-1\right)^3=-1\)

d)  \(D=x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=\left(-1\right)^3-3.\left(-12\right).\left(-1\right)=-37\)

          \(4x^2+y^2-4x+10y+26=0\)

\(\Leftrightarrow\)\(\left(4x^2-4x+1\right)+\left(y^2+10y+25\right)=0\)

\(\Leftrightarrow\)\(\left(2x-1\right)^2+\left(y+5\right)^2=0\)

\(\Leftrightarrow\)\(\hept{\begin{cases}2x-1=0\\y+5=0\end{cases}}\)

\(\Leftrightarrow\)\(\hept{\begin{cases}x=\frac{1}{2}\\y=-5\end{cases}}\)

Vậy..

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

\(=x^4x^2+1.x^4-\left(x^2\right)^2+1.x^2-1.x^2+1.1\)

\(=x^6+x^4-x^4+1\)

\(=x^6+1\)

bạn áp dụng HĐT:  \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)

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

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

\(=x^6+1\)

x^2+y^2-x+6y+10=x(x-1)+y(y+6)+10

=>CTNN của biểu thức=10 <=>x=0;y=0

Gọi bt là A, ta có:

\(A=x^2+y^2-x+6y+10=\left(x^2-2.\frac{1}{2}.x+\frac{1}{4}\right)+\left(y^2+2.\frac{1}{2}.x+\frac{1}{4}\right)\)

Ta xét: \(\left(x-\frac{1}{2}\right)^2\ge0\) (bình phương lên)

           \(\left(y-3\right)^2\ge0\) (bình phương lên)

\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\left(y-3\right)^2\ge0\)

\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\left(y-3\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

\("="\Leftrightarrow x=\frac{1}{2};x=3\)