Dap an la : -2, -4,-6

g/ x(x-2)-x²=5x-7

\(\Leftrightarrow x^2-2x-x^2=5x-7\\ \Leftrightarrow7x=7\Leftrightarrow x=1\)

h/\(3x\left(x-7\right)+2\left(x-7\right)=0\\ \Leftrightarrow3x^2-19x-14=0\\ \)

\(\Leftrightarrow\left(x-7\right)\left(3x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\frac{2}{3}\end{matrix}\right.\)

cac giup minh di minh sap phai nop roi

a²+4b²⁺4c²>= 4ab-4ac+8bc

a²+4b²⁺4c² - 4ab +4ac-8bc

(a² - 4ab+4b²)+4c²+(4ac-8bc>=0)

suy ra (a-2b²)+2.2c.(a-2b)+(2c)²

(a-2b+2c)²>=0

dau = xảy ra khi va chỉ khi a+2c=2b

a²+4b²+4c²>= 4ab-4ac+8bc(dpcm)

ban giai day du cho minh di. minh lam de nop ma

bó tay 

bá tay luon,cá khi bá nốt chan

\(a^2+ab+b^2=a^2+\frac{2.a.1}{2}b+\frac{1}{4}b^2+\frac{3}{4}b^2=\left(a+\frac{1}{2}b\right)^2+\frac{3}{4}b^2\ge0\)

\(\left(a^2+b^2\right)^2\ge ab.\left(a+b\right)^2\)

\(\Leftrightarrow a^4+2a^2b^2+b^4\ge ab.\left(a^2+2ab+b^2\right)\)

\(\Leftrightarrow a^4+2a^2b^2+b^4-a^3b-2a^2b^2-ab^3\ge0\)

\(\Leftrightarrow a^4-a^3b+b^4-ab^3\ge0\)

\(\Leftrightarrow a^{\text{3}}.\left(a-b\right)-b^3.\left(a-b\right)\ge0\)

\(\Leftrightarrow\left(a-b\right).\left(a-b\right).\left(a^2-ab+b^2\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)^2.\left(a^2-ab+b^2\right)\ge0\text{ vì }\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(a^2-ab+b^2\right)\ge0\left(cmt\right)\end{cases}}\)

Vì BĐT cuối đúng nên BĐT đầu đúng (đpcm) 

2.x - 12.x = 0 

=> 2 - 12 . x = 0 

=> - 10.x     = 0 

=>       x      = 0 : ( - 10 ) 

=>       x      = 0 

???? 

2xX-12xX=0

<=>x(2-12)=0

<=>x=0

\(2\times x-12\times x=0\)

   \(\left(2-12\right)\times x=0\) 

             \(-10\times x=0\)

                             \(x=0:\left(-10\right)\)

                               \(x=0\)

Vậy x = 0

X - 45 = 23

X = 23 + 45

X = 68

ĐÚNG ĐÓ

x - 45 = 23

x= 23+45 

x= 68

X-45=23

X = 23+45

X=68

\(M=1+\dfrac{1}{5}+\dfrac{3}{35}+...+\dfrac{3}{9999}\\ =\dfrac{3}{3}+\dfrac{3}{15}+\dfrac{3}{35}+...+\dfrac{3}{9999}\\ =\dfrac{3}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\\ =\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\\ =\dfrac{3}{2}\left(1-\dfrac{1}{101}\right)=\dfrac{3}{2}\cdot\dfrac{100}{101}=\dfrac{150}{101}\)

\(\frac{x^3-x^2-x-2}{x^5-3x^4+4x^3-5x^2+3x-2}\)

\(=\frac{x^3-2x^2+x^2-2x+x-2}{x^5-2x^4-x^4+2x^3+2x^3-4x^2-x^2+2x+x-2}\)

\(=\frac{\left(x^3-2x^2\right)+\left(x^2-2x\right)+\left(x-2\right)}{\left(x^5-2x^4\right)-\left(x^4-2x^3\right)+\left(2x^3-4x^2\right)-\left(x^2-2x\right)+\left(x-2\right)}\)

\(=\frac{x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)}{x^4\left(x-2\right)-x^3\left(x-2\right)+2x^2\left(x-2\right)-x\left(x-2\right)+\left(x-2\right)}\)

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