Đường thẳng song song d nên nhận (3;-4) là 1 vtpt

Phương trình:

\(3\left(x-2\right)-4\left(y-1\right)=0\Leftrightarrow3x-4y-2=0\)

a: =>3M+2x^4y^4=x^4y^4

=>3M=-x^4y^4

=>M=-1/3*x^4y^4

b: x^2-2M=3x^2

=>2M=-2x^2

=>M=-x^2

c: =>M=-x^2y^3-3x^2y^3=-4x^2y^3

d: =>M=7x^2y^2-3x^2y^2=4x^2y^2

\(x=\frac{7+4y}{3}\Rightarrow3x^2+4y^2=3.\left(\frac{7+4y}{3}\right)^2+4y^2=\frac{\left(7+4y\right)^2}{3}+4y^2\)

\(=\frac{49+56y+16y^2+12y^2}{3}=\frac{49+56y+28y^2}{3}\)

\(=\frac{28.\left(\frac{7}{4}+2y+y^2\right)}{3}=\frac{28.\left(y^2+2y+1+\frac{3}{4}\right)}{3}=\frac{28\left(y+1\right)^2+21}{3}\)

\(\ge\frac{21}{3}=7\)

cách dễ nhất tính x theo y thế số vô làm

Lời giải:

a. $=(2x)^2-2.2x.5y+(5y)^2=4x^2-20xy+25y^2$
b. $=(3x)^2+2.3x.2y+(2y)^2=9x^2+12xy+4y^2$

c. $=(4y+3x)(4y-3x)=(4y)^2-(3x)^2=16y^2-9x^2$

\(a.4x^2-10xy+25y^2\)

\(b.9x^2+6xy+4y^2\)

\(c.16y^2-9x^2\)

đáp án A là đáp án đúng

B là câu trả lời đug đó bạn

Ta có: \(3x-4y=7\) \(\Rightarrow x=\dfrac{7+4y}{3}\)

Thay vào ta được:

\(3.\left(\dfrac{7+4y}{3}\right)^2+4y^2=3.\dfrac{49+56y+16y^2}{9}+4y^2\)

\(=\dfrac{147+168y+48y^2+36y^2}{9}=\dfrac{84y^2+168y+147}{9}=\dfrac{84\left(y^2+2y+\dfrac{7}{4}\right)}{9}=\dfrac{84\left(y+1\right)^2+63}{9}\ge\dfrac{63}{9}=7\)⇒ ĐPCM

A) \(-4x2xy^2+3x^2.\frac{1}{3}y+\left(-5\right)xy.\frac{1}{5}xy=-8x^2y^2+x^2y+\left(-x^2y^2\right)=-9x^2y^2+x^2y\)

B) \(\frac{4}{3}x^4y^7-3x^4y^7=\frac{-5}{3}x^4y^7\)

C) \(\frac{2}{3}x^3y^4+3x^3y^4=3\frac{2}{3}x^3y^4\)

CHÚC BN HỌC TỐT!!!

3x=4y

nên x/4=y/3

Đặt x/4=y/3=k

=>x=4k; y=3k

\(H=\dfrac{2xy+3x^2}{3xy+4y^2}=\dfrac{2\cdot4k\cdot3k+3\cdot16k^2}{3\cdot4k\cdot3k+4\cdot9k^2}\)

\(=\dfrac{24k^2+48k^2}{36k^2+36k^2}=1\)

3x²+4y²=7xy

<=> 3x²-3xy+4y²-4xy=0

<=> 3x(x-y)-4y(x-y)=0

<=> (3x-4y)(x-y)=0

<=> 3x-4y=0 hoặc x-y=0

<=> 3x=4y hoặc x=y

<=> y = \(\frac{3}{4}\)x hoặc x=y

+) y = \(\frac{3}{4}\)x, ta có:

F = \(\frac{4.\frac{3}{4}x+2x}{5.\frac{3}{4}x-7x}+\)\(\frac{3x-2.\frac{3}{4}x}{10.\frac{3}{4}x-4x}\)

F = \(\frac{5x}{-\frac{13}{4}x}+\frac{\frac{3}{2}x}{\frac{7}{2}x}\)

F = \(-\frac{20}{13}+\frac{3}{7}=-\frac{101}{91}\)

+) x = y, ta có:

F = \(\frac{4x+2x}{5x-7x}+\frac{3x-2x}{10x-4x}\)

F = \(\frac{6x}{-2x}+\frac{1x}{6x}=-3+\frac{1}{6}=-\frac{17}{6}\)

Từ \(3x^2+4y^2=7xy\Rightarrow3x^2+4y^2-7xy=0\)

\(\Rightarrow3x^2-4xy-3xy+4y^2=0\)

\(\Rightarrow x\left(3x-4y\right)-y\left(3x-4y\right)=0\)

\(\Rightarrow\left(x-y\right)\left(3x-4y\right)=0\)\(\Rightarrow\left[\begin{matrix}x=y\\x=\frac{4y}{3}\end{matrix}\right.\)

*)Xét \(x=y\) ta có \(F=\frac{4y+2y}{5y-7y}+\frac{3y-2y}{10y-4y}=\frac{6y}{-2y}+\frac{y}{6y}=-3+\frac{1}{6}=-\frac{17}{6}\)

*)Xét \(x=\frac{4y}{3}\) ta có \(F=\frac{4y+2\cdot\frac{4y}{3}}{5y-7\cdot\frac{4y}{3}}+\frac{3\cdot\frac{4y}{3}-2y}{10y-4\cdot\frac{4y}{3}}=\frac{4y+\frac{8y}{3}}{5y-\frac{28y}{3}}+\frac{4y-2y}{10y-\frac{16y}{3}}=\frac{-20}{13}+\frac{3}{7}=\frac{-101}{91}\)