Lời giải:

$A=2x^2+3y^2-2xy+4x-7y+2012$

$2A=4x^2+6y^2-4xy+8x-14y+4024$

$=(4x^2-4xy+y^2)+5y^2+8x-14y+4024$

$=(2x-y)^2+4(2x-y)+5y^2-10y+4024$

$=(2x-y)^2+4(2x-y)+4+5(y^2-2y+1)+4015$

$=(2x-y+2)^2+5(y-1)^2+4015\geq 4015$

$\Rightarrow A\geq \frac{4015}{2}$

Vậy $A_{\min}=\frac{4015}{2}$ khi $(x,y)=(\frac{-1}{2},1)$

1) \(4x=7y\Leftrightarrow\dfrac{x}{7}=\dfrac{y}{4}\Rightarrow\dfrac{x^2}{49}=\dfrac{y^2}{16}=\dfrac{x^2+y^2}{49+16}=\dfrac{260}{65}=4\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2=4.49=196\\y^2=4.16=64\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=14,y=8\\x=-14,y=-8\end{matrix}\right.\) (vì \(\dfrac{x}{7}=\dfrac{y}{4}\) nên \(x,y\) cùng dấu) 

2) \(2^{x-1}+5.2^{x-2}=\dfrac{7}{32}\)

\(\Leftrightarrow2^{x-1}+\dfrac{5}{2}.2^{x-1}=\dfrac{7}{32}\)

\(\Leftrightarrow2^{x-1}=\dfrac{1}{16}=2^{-4}\)

\(\Leftrightarrow x-1=-4\)

\(\Leftrightarrow x=-3\)

3) \(\left|x+5\right|+\left(3y-4\right)^{2016}=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+5=0\\3y-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-5\\y=\dfrac{4}{3}\end{matrix}\right.\)

VV

1D  2C

Câu 1: D

Câu 2: C

đề bài nghĩa là sao hả bạn ????????????????
 

ta có: \(4x=3y\Rightarrow\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{20}.\)

\(7y=5z\Rightarrow\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{y}{20}=\frac{z}{28}\)

\(\Rightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{28}\)

ADTCDTSBN

có: \(\frac{x}{15}=\frac{y}{20}=\frac{x-y}{15-20}=\frac{-46}{-5}=\frac{46}{5}\)

...

ta có: \(4x=3y\Rightarrow\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{20}.\)

\(7y=5z\Rightarrow\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{y}{20}=\frac{z}{28}\)

\(\Rightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{28}\)

ADTCDTSBN

có: \(\frac{x}{15}=\frac{y}{20}=\frac{x-y}{15-20}=\frac{-46}{-5}=\frac{46}{5}\)

...

#

a) Vì \(\hept{\begin{cases}\left|5-4x\right|\ge0\\\left|7y-3\right|\ge0\end{cases}}\)nên dấu "=" xảy ra <=> x = 5/4 ; y = 3/7

b) Vì \(\hept{\begin{cases}\left|x-3y-1\right|\ge0\\\left|y-4\right|\ge0\end{cases}}\)nên dấu "=" xảy ra <=> x = 13 ; y = 4

a)do |5-4x|+|7y-3|=0,mà|5-4x| và|7y-3| đều lớn hơn hoặc = 0

suy ra 5-4x=7y-3=0 thì biểu thức mới thỏa mãn

(do mọi số trong  dấu GTTĐ đều lớn hơn hoặc bằng 0)

tự giải nốt nhé

a.

\(\left(x^2-x+1\right)\left(x^2-x+2\right)=12\)

Đặt \(x^2-x+1=y\) ta được:

\(y\left(y+1\right)=12\)

\(\Leftrightarrow y^2+y-12=0\)

\(\Leftrightarrow y^2+4y-3y-12=0\)

\(\Leftrightarrow\left(y-3\right)\left(y+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=3\\y=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+1=3\\x^2-x+1=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-x-2=0\\x^2-x+5=0\left(vn\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)

b.

\(3y^3-7y^2-7y+3=0\)

\(\Leftrightarrow3\left(y^3+1\right)-7y\left(y+1\right)=0\)

\(\Leftrightarrow3\left(y+1\right)\left(y^2-y+1\right)-7y\left(y+1\right)=0\)

\(\Leftrightarrow\left(y+1\right)\left(3y^2-3y+3-7y\right)=0\)

\(\Leftrightarrow\left(y+1\right)\left(3y^2-10y+3\right)=0\)

\(\Leftrightarrow\left(y+1\right)\left(3y-1\right)\left(y-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=-1\\y=\dfrac{1}{3}\\y=3\end{matrix}\right.\)

c.

\(x^2+2y^2=4x+4y-6=0\)

\(\Leftrightarrow x^2-4x+4+2y^2-4y+2=0\)

\(\Leftrightarrow\left(x-2\right)^2+2\left(y-1\right)^2=0\)

Do \(\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\\\left(y-1\right)^2\ge0\end{matrix}\right.\) với mọi x;y

\(\Rightarrow\left(x-2\right)^2+2\left(y-1\right)^2\ge0\) ; \(\forall x;y\)

Dấu "=" xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}x-2=0\\y-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

\(\frac{1+3y}{12}=\frac{1+3y}{5x}=\frac{1+7y}{4x}\)(1)

Từ \(\frac{1+3y}{12}=\frac{1+3y}{5x}\)\(\Rightarrow5x=12\)\(\Rightarrow x=\frac{12}{5}\)

Thay \(x=\frac{12}{5}\)vào (1) ta có: \(\frac{1+3y}{12}=\frac{1+7y}{4.\frac{12}{5}}=\frac{1+7y}{\frac{48}{5}}=\frac{5\left(1+7y\right)}{48}\)

\(\Rightarrow\frac{4\left(1+3y\right)}{48}=\frac{5\left(1+7y\right)}{48}\)

\(\Rightarrow4\left(1+3y\right)=5\left(1+7y\right)\)\(\Leftrightarrow4+12y=5+35y\)

\(\Leftrightarrow23y=-1\)\(\Leftrightarrow y=-\frac{1}{23}\)

Vậy \(x=\frac{12}{5}\)và \(y=-\frac{1}{23}\)

...có đáp án mik đã đến muộn 6 năm