Đáp án:
Bài 20:
a.$ x=\dfrac1{10}$ hoặc $x=-\dfrac1{10}$
b.$x=3$ hoặc $x=-1$
c.$x=\dfrac{13}{15}$
Giải thích các bước giải:
Bài 20:
a.Ta có:
$|\dfrac53x|=|-\dfrac16|$
$\to \dfrac53|x|=\dfrac16$
$\to |x|=\dfrac1{10}$
$\to x=\dfrac1{10}$ hoặc $x=-\dfrac1{10}$
b.Ta có:
$|\dfrac34x-\dfrac34|-\dfrac34=|-\dfrac34|$
$\to|\dfrac34x-\dfrac34|-\dfrac34=\dfrac34$
$\to|\dfrac34(x-1)|=\dfrac32$
$\to \dfrac34|x-1|=\dfrac32$
$\to |x-1|=2$
$\to x-1=2\to x=3$ hoặc $x-1=-2\to x=-1$
c.Ta có:
$|x+\dfrac35|-|x-\dfrac73|=0$
$\to |x+\dfrac35|=|x-\dfrac73|$
$\to x+\dfrac35=x-\dfrac73\to \dfrac35=-\dfrac73$ vô lý
Hoặc $x+\dfrac35=-(x-\dfrac73)$
$\to x+\dfrac35=-x+\dfrac73$
$\to 2x=\dfrac{26}{15}$
$\to x=\dfrac{13}{15}$
Bài 21:
a.Ta có:
$(x+y)(x+y)=x(x+y)+y(x+y)=x^2+xy+yx+y^2=x^2+2xy+y^2$
b.Ta có:
$(x-y)(x-y)=x(x-y)-y(x-y)=x^2-xy-yx+y^2=x^2-2xy+y^2$
c.Ta có:
$(x+y)(x-y)=x(x-y)+y(x-y)=x^2-xy+yx-y^2=x^2-y^2$
d.Ta có:
$(x+5)(x-1)=x(x-1)+5(x-1)=x^2-x+5x-5=x^2+4x-5$
