Đại số lớp 7

a, \(\left(a+b\right)^2=\left(a+b\right)\left(a+b\right)=a^2+ab+ab+b^2=a^2+2ab+b^2\)

b, \(\left(a-b\right)^2=\left(a-b\right)\left(a-b\right)=a^2-ab-ab+b^2=a^2-2ab+b^2\)

c, \(a^2-b^2=a^2+ab-ab-b^2=a\left(a+b\right)-b\left(a+b\right)=\left(a+b\right)\left(a-b\right)\)

\(a,\left(a+b\right)^2\\ =\left(a+b\right)\left(a+b\right)\\ =a^2+ab+b^2+ab\\ =a^2+2ab+b^2.\)

\(b,\left(a-b\right)^2\\ =\left(a-b\right)\left(a-b\right)\\ =a^2-ab+b^2-ab\\ =a^2-2ab+b^2.\)

\(c,a^2-b^2\\ =a^2+ab-ab-b^2\\ =a\left(a+b\right)-b\left(a+b\right)\\ =\left(a-b\right)\left(a+b\right).\)

tìm gtnn hả bạn ? 

\(A=\left(2x-1\right)^{2000}+48\ge48\)

Dấu ''='' xảy ra khi x = 1/2 

\(B=\left(x-5\right)^{200}+3\left(y-4\right)^{302}+67\ge67\)

Dấu ''='' xảy ra khi x = 5 ; y = 4 

Theo tc dãy tỉ số bằng nhau ta có 

\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{6}=\dfrac{x+y+z}{3+4+6}=\dfrac{52}{13}=4\Rightarrow x=12;y=16;z=24\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

`x/3 = y/4 = z/6 = (x+y+z)/(3+4+6) = 52/13 =4`

`x/3 =4=>x=3.4=12

`y/4 = 4=>y=4.4=16`

`z/6 =4=>z=4.6=24`

\(Taco:\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{6}=\dfrac{x+y+z}{3+4+6}=\dfrac{52}{13}=4\\ =>x=3.4=12\\ y=4.4=16\\ z=6.4=24\)

Ta có x^2/9 = y^2 / 49 

Theo tc dãy tỉ số bằng nhau ta có 

\(\dfrac{x^2}{9}=\dfrac{y^2}{49}=\dfrac{x^2-y^2}{9-49}=-\dfrac{360}{-40}=9\Rightarrow x=9;y=21\)

4x=5y nên x/5=y/4

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{2x-5y}{2\cdot5-5\cdot4}=\dfrac{40}{-10}=-4\)

Do đó: x=-20; y=-16

x/5=y/4 Theo tc dãy tỉ số bằng nhau ta có 

\(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{2x-5y}{10-20}=\dfrac{40}{-10}=-4\Rightarrow x=-20;y=-16\)

Lời giải:

$4x=5y\Rightarrow \frac{x}{5}=\frac{y}{4}$

$\Rightarrow \frac{2x}{10}=\frac{5y}{20}$

Áp dụng TCDTSBN:

$\frac{2x}{10}=\frac{5y}{20}=\frac{2x-5y}{10-20}=\frac{40}{-10}=-4$

$\Rightarrow x=-4.10:2=-20; y=-4.20:5=-16$

x/6=y/5 Theo tc dãy tỉ số bằng nhau ta có 

\(\dfrac{x}{6}=\dfrac{y}{5}=\dfrac{x+y}{6+5}=\dfrac{121}{11}=11\Rightarrow x=66;y=55\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{6}=\dfrac{y}{5}=\dfrac{x+y}{6+5}=\dfrac{121}{11}=11\)

Do đó: x=66; y=55

`|x-1|-2x=1/2`

`|x-1|=2x+1/2`

TH1: `x-1>=0`

`x-1=2x+1/2`

`-x=3/2`

`x=-3/2` (Loại)

TH2: `x-1<0`

`-(x-1)=2x+1/2`

`-x+1=2x+1/2`

`-3x=-1/2`

`x=1/6` (TM)

Vậy `x=1/6`.

`|-2x + 1,5| = 1/4`

`=>` \(\left[{}\begin{matrix}-2x+1,5=\dfrac{1}{4}\\-2x+1,5=-\dfrac{1}{4}\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}-2x=\dfrac{1}{4}-1,5\\-2x=-\dfrac{1}{4}-1,5\end{matrix}\right.\)

`=>` \(\left[{}\begin{matrix}-2x=-\dfrac{5}{4}\\-2x=-\dfrac{7}{4}\end{matrix}\right.\)

`=>` \(\left[{}\begin{matrix}x=-\dfrac{5}{4}:\left(-2\right)\\x=-\dfrac{7}{4}:\left(-2\right)\end{matrix}\right.\)

`=>` \(\left[{}\begin{matrix}x=-\dfrac{5}{4}.\dfrac{-1}{2}\\x=-\dfrac{7}{4}.\dfrac{-1}{2}\end{matrix}\right.\)

`=>` \(\left[{}\begin{matrix}x=\dfrac{5}{8}\\x=\dfrac{7}{8}\end{matrix}\right.\)

`#H.J`

`|-2x + 1,5| = 1/4`

`=> |-2x + 1,5| = 0,25`

`=>` \(\left[{}\begin{matrix}-2x+1,5=0,25\\-2x+1,5=-0,25\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}-2x=0,25-1,5\\-2x=-0,25-1,5\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}-2x=-1,25\\-2x=-1,75\end{matrix}\right.\)

`=>` \(\left[{}\begin{matrix}x=0,625\\x=0,875\end{matrix}\right.\)

`2/3 + 7/4 : x = 5/6`

`7/4 : x = 5/6 - 2/3`

`7/4 : x = 5/6 - 4/6`

`7/4 : x =1/6`

`x=7/4:1/6`

`x=7/4xx6`

`x=21/2`

`(x-1/3).(x+2/5)=0`

`=>` \(\left[{}\begin{matrix}x-\dfrac{1}{3}=0\\x+\dfrac{2}{5}=0\end{matrix}\right.\)

`=>` \(\left[{}\begin{matrix}x=0+\dfrac{1}{3}\\x=0-\dfrac{2}{5}\end{matrix}\right.\)

`=>` \(\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{2}{5}\end{matrix}\right.\)

Vậy.....

`#H.J`