Bạn là Lovemath bên Hoidap247 hã ?

$3x=4y\to \dfrac x 4=\dfrac  y 3$

$4y=5z\to \dfrac y 5=\dfrac z 4$

$\to \dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{12}$

Áp dụng tính chất dãy tỉ số bằng nhau

$\dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x-(y+z)}{20-(15+12)}=\dfrac{-21}{-7}=3$

$\to \begin{cases}x=60\\y=45\\z=36\end{cases}$

Vậy $(x;y;z)=(60;45;36)$

l/ $6x^2(x-1)-9x(x-1)\\=(6x^2-9)(x-1)\\=3(2x^2-3)(x-1)\\=3(\sqrt2 x-\sqrt 3)(\sqrt 2 x+\sqrt 3)(x-1)$

m/ $4x^2(x-2)+9x(2-x)\\=4x^2(x-2)-9x(x-2)\\=(4x^2-9x)(x-2)\\=x(4x-9)(x-2)$

n/ $4x^2y-4xy+y\\=y(4x^2-4x+1)\\=y(2x-1)^2$

o/ $3x(2x-3y)-6(3y-2x)\\=3x(2x-3y)+6(2x-3y)\\=(3x+6)(2x-3y)\\=3(x+2)(2x-3y)$

p/ $4x^2(x-1)+(1-x)\\=4x^2(x-1)-(x-1)\\=(4x^2-1)(x-1)\\=(2x-1)(2x+1)(x-1)$

$\sin a.\cos a.(\tan a+\cot a)\\=\sin a.\cos a.\tan a+\sin a.\cos a.\cot a\\=\sin a.\cos a.\dfrac{\sin a}{\cos a}+\sin a.\cos a.\dfrac{\cos a}{\sin a}\\=\sin^2 a+\cos^2 a\\=1$

(4x^2)/3+2x-1/27

20-[30(5-1)²]

=20-30.4²

=20-30.16

=20-480

=-460

(1):chuyển động

(2):đứng yên

Bài 5.

Ta có: $(a-b)^2+(a-c)^2+(b-c)^2\ge 0$

`<=>a^2-2ab+b^2+a^2-2ac+c^2+b^2-2bc+c^2>=0`

`<=>2a^2+2b^2+2c^2>=2ab+2ac+2bc`

`<=>2(a^2+b^2+c^2)>= 2(ab+ac+bc)`

`2ab+2ac+2bc`

`=ab+ab+ac+ac+bc+bc`

`=a(b+c)+b(a+c)+c(a+b)`

Vì `a+b+c=`

$\Rightarrow \begin{cases}a+b=2-c\\b+c=2-a\\a+c=2-b\end{cases}$

Thay $a+b=2-c,b+c=2-a,a+c=2-b$ vào $a(b+c)+b(a+c)+c(a+b)$

$a(2-a)+b(2-b)+c(2-c)\\=2a+2b+2c-(a^2+b^2+c^2)\\=2(a+b+c)-(a^2+b^2+c^2)\\=4-(a^2+b^2+c^2)$

mà $2(a^2+b^2+c^2)\ge 2(ab+ac+bc)$

$\RIghtarrow 2(a^2+b^2+c^2)\ge 4-(a^2+b^2+c^2)\\\Leftrightarrow 3(a^2+b^2+c^2)\ge 4\\\Leftrightarrow a^2+b^2+c^2\ge \dfrac{4}{3}$

$\Rightarrow$ Dấu "=" xảy ra khi $a=b=c$

mà $a+b+c=2$

$\to a=b=c=\dfrac{2}{3}$

Vậy $A$ đạt GTNN là `4/3` tại `a=b=c=2/3`

$4a=7b\Leftrightarrow \dfrac{a}{7}=\dfrac{b}{4}$

$\Leftrightarrow \dfrac{a^2}{49}=\dfrac{b^2}{16}$

Áp dụng tính chất dãy tỉ số bằng nhau:

$\dfrac{a^2}{49}=\dfrac{b^2}{16}=\dfrac{a^2+b^2}{49+16}=\dfrac{260}{65}=4$

$\Rightarrow \begin{cases}\dfrac{a^2}{49}=4\\\dfrac{b^2}{16}=4\end{cases}$

$\Leftrightarrow\begin{cases}a^2=196\\b^2=64\end{cases}$

$\Leftrightarrow \begin{cases}a=\pm 14\\b=\pm 8\end{cases}$

Vậy $a=\pm 14;b=\pm 8$

$z=\dfrac{-8.48}{-16}=24$ nhé