Toán Lớp 8: tim gia tri nho nhat cua cac bieu thuc A=(x+1)(2x-1),B=4x^2-4xy+2y^2+1

Question

Toán Lớp 8:  tim gia tri nho nhat cua cac bieu thuc A=(x+1)(2x-1),B=4x^2-4xy+2y^2+1

quynhmaithu · Answer

$A ,=(x+1)(2x-1)   quad =2x^2+x-1   quad =2 left(x^2+ dfrac{1}{2}x- dfrac{1}{2} right)   quad=2 left(x^2+2.x. dfrac{1}{4}+ dfrac{1}{16}- dfrac{9}{16} right)   quad=2 left(x+ dfrac{1}{4} right)^2- dfrac{9}{8}$   Nhận thấy: $ left(x+ dfrac{1}{4} right)^2 ge 0$   $&harr;2 left(x+ dfrac{1}{4} right)^2 ge 0  &harr;2 left(x+ dfrac{1}{4} right)^2- dfrac{9}{8} ge - dfrac{9}{8}$   $&rarr; min A=- dfrac{9}{8}$   $&rarr;$ Dấu '=' xảy ra khi $x+ dfrac{1}{4}=0$   $&harr;x=- dfrac{1}{4}$   Vậy $ min A=- dfrac{9}{8}$ với $x=- dfrac{1}{4}$   $B ,=4x^2-4xy+2y^2+1   quad =4x^2-4xy+y^2+y^2+1   quad =(4x^2-4xy+y^2)+y^2+1   quad =(2x-y)^2+y^2+1$   Nhận thấy: $ begin{cases}(2x-y)^2 ge 0  y^2 ge 0 end{cases}$   $&harr;(2x-y)^2+y^2 ge 0  &harr;(2x-y)^2+y^2+1 ge 1  &rarr; min B=1$   $&rarr;$ Dấu '=' xảy ra khi $ begin{cases}2x-y=0  y=0 end{cases}$   $&harr; begin{cases}2x=0  y=0 end{cases}  &harr; begin{cases}x=0  y=0 end{cases}  &harr;x=y=0$   Vậy $ min B=1$ khi $x=y=0$

tonhitruc · Answer

qquad A=(x+1)(2x-1) A=2x^2-x+2x-1 A=2x^2+x-1 A=2(x^2+1/2x-1/2) A=2(x^2+2.x. 1/4+1/16-9/16) A=2(x+1/4)^2-9/8>=-9/8 Dấu = xảy ra khi x+1/4=0<=>x=-1/4 Vậy A_(min)=-9/8<=>x=-1/4 qquad B=4x^2-4xy+2y^2+1 B=(4x^2-4xy+y^2)+y^2+1 B=(2x-y)^2+y^2+1>=1 Dấu = xảy ra khi {(2x-y=0),(y=0):}<=>x=y=0 Vậy B_(min)=1<=>x=y=0