Hằng đẳng thức đáng nhớ(Cần gấp)

T

trinhvy_123

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E

eunhyuk_0330

Bài 1:
a) $6x-x^2-9=-(x^2-6x+9)=-(x-3)^2$
Do $(x-3)^2$ \geq 0 \Rightarrow$-(x-3)^2$\leq với mọi x
b) $10x-25-x^2=-(x^2-10x+25)=-(x-5)^2$\leq 0
c) $-x^2-4x-10=-(x^2+4x+4)-6=-(x+2)^2-6$\leq -6 <0
d)$12x-9x^2-7=-(9x^2-12x+4)-3=-(3x-2)^2-3$\leq -3<0
 
E

eunhyuk_0330

Bài 2:
a) $x^2-10x+26+y^2+2y=0$
\Leftrightarrow $(x^2-10x+25)+(y^2+2y+1)=0$
\Leftrightarrow $(x-5)^2+(y+1)^2=0$
\Leftrightarrow $x-5=y+1=0$
\Leftrightarrow $x=5;y=-1$
b) $(2x+5)^2-(x-7)^2=0$
\Leftrightarrow $(2x+5+x-7)(2x+5-x+7)=0$

\Leftrightarrow $(3x-2)(x+12)=0$
\Leftrightarrow $3x-2=0 hoặc x+12=0$
\Leftrightarrow $x=\dfrac{3}{2} hoặc x=-12$
c) $25(x-3)^2=49(1-2x)^2$
\Leftrightarrow $[5(x-3)]^2=[7(1-2x)]^2$
\Leftrightarrow $(5x-15)^2-(7-14x)^2=0$
\Leftrightarrow $(5x-15+7-14x)(5x-15-7+14x)=0$
\Leftrightarrow $(-9x-8)(19x-22)=0$
\Leftrightarrow $-9x-8=0 hoặc 19x-22=0$
\Leftrightarrow $x=\dfrac{-8}{9} hoặc x=\dfrac{22}{19}$
 
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