If `p/a+q/b+r/c=1` and `a/p+b/q+c/r=0` then find the value of the expression `T=p^2/a^2+q^2/b^2+r^2/c^2` 11/10/2021 Bởi Clara If `p/a+q/b+r/c=1` and `a/p+b/q+c/r=0` then find the value of the expression `T=p^2/a^2+q^2/b^2+r^2/c^2`
Đáp án: $+)$$\frac{p}{a}+\frac{q}{b}+\frac{r}{c}=1$ $(\frac{p}{a}+\frac{q}{b}+\frac{r}{c})^2=1$ $\frac{p^2}{a^2}+\frac{q^2}{b^2}+\frac{r^2}{c^2}+2(\frac{pq}{ab}+\frac{qr}{bc}+\frac{pr}{ac}) =1$ $T+2(\frac{pqc+qra+prb}{abc}) =1$ (1) Other face: $\frac{a}{p}+\frac{b}{q}+\frac{c}{r}=0$ $⇔\frac{pqc+qra+prb}{pqr}=0$ $⇔pqc+qra+prb=0 $(2) $(1),(2)⇒T=1$ Bình luận
Đáp án: Giải thích các bước giải: Let pa=x,qb=y,rc=z.pa=x,qb=y,rc=z. So given equations are: x+y+z=1x+y+z=1 —(1) 1x+1y+1z=01x+1y+1z=0 —(2) Required: x2+y2+z2x2+y2+z2 From (2), yz+xz+xyxyz=0yz+xz+xyxyz=0 => yz+xz+xy=0yz+xz+xy=0 —(3) Squaring (1) on both sides: x2+y2+z2+2(xy+yz+xz)=1x2+y2+z2+2(xy+yz+xz)=1 => x2+y2+z2+2(0)=1x2+y2+z2+2(0)=1 ∵yz+xz+xy=0∵yz+xz+xy=0 => x2+y2+z2=1 Bình luận
Đáp án:
$+)$$\frac{p}{a}+\frac{q}{b}+\frac{r}{c}=1$
$(\frac{p}{a}+\frac{q}{b}+\frac{r}{c})^2=1$
$\frac{p^2}{a^2}+\frac{q^2}{b^2}+\frac{r^2}{c^2}+2(\frac{pq}{ab}+\frac{qr}{bc}+\frac{pr}{ac}) =1$
$T+2(\frac{pqc+qra+prb}{abc}) =1$ (1)
Other face: $\frac{a}{p}+\frac{b}{q}+\frac{c}{r}=0$
$⇔\frac{pqc+qra+prb}{pqr}=0$
$⇔pqc+qra+prb=0 $(2)
$(1),(2)⇒T=1$
Đáp án:
Giải thích các bước giải:
Let pa=x,qb=y,rc=z.pa=x,qb=y,rc=z. So given equations are:
x+y+z=1x+y+z=1 —(1)
1x+1y+1z=01x+1y+1z=0 —(2)
Required: x2+y2+z2x2+y2+z2
From (2), yz+xz+xyxyz=0yz+xz+xyxyz=0
=> yz+xz+xy=0yz+xz+xy=0 —(3)
Squaring (1) on both sides:
x2+y2+z2+2(xy+yz+xz)=1x2+y2+z2+2(xy+yz+xz)=1
=> x2+y2+z2+2(0)=1x2+y2+z2+2(0)=1 ∵yz+xz+xy=0∵yz+xz+xy=0
=> x2+y2+z2=1