Nếu x < -2

=> |x + 2| = -(x + 2) = -x - 2

=> |x - 5| = -(x - 5) = -x + 5

Khi đó |x + 2| + |x - 5| = 3x (1)

<=> -x - 2 - x + 5 = 3x

=> 3 = 5x

=> x = 0,6 (loại)

Nếu \(-2\le x\le5\)

=> |x + 2| = x + 2 

=> |x - 5| = -(x - 5) = -x + 5

Khi đó (1) <=> x + 2 - x + 5 = 3x

=> 3x = 7

=> x = 7/3 (tm)

Nếu x > 5

=> |x + 2| = x + 2

=> |x - 5| = x - 5

Khi đó (1) <=> x + 2 + x - 5 = 3x

=> 2x - 3 = 3x

=> x = -3 (loại) 

Vậy x = 7/3

| x + 2 | ≥ 0 <=> x + 2 ≥ 0 => x ≥ -2

| x - 5 | ≥ 0 <=> x - 5 ≥ 0 => x ≥ 5

Vậy để giải phương trình trên ta xét ba trường hợp

1/ x < -2

Pt trở thành : 

-( x + 2 ) - ( x - 5 ) = 3x

<=> -x - 2 - x + 5 = 3x

<=> -2x + 3 = 3x

<=> -2x - 3x = -3

<=> -5x = -3

<=> x = 3/5 ( không tmđk )

2/ -2 < x < 5

Pt trở thành

( x + 2 ) - ( x - 5 ) = 3x

<=> x + 2 - x + 5 = 3x

<=> 7 = 3x

<=> x = 7/3 ( tmđk )

3/ x ≥ 5

Pt trở thành :

x + 2 + x - 5 = 3x

<=> 2x - 3 = 3x

<=> 2x - 3x = 3

<=> -x = 3

<=> x = -3 ( không tmđk )

Vậy phương trình có nghiệm duy nhất là x = 7/3

Gọi \(P\left(x\right)=ax^4+bx^3+cx^2+dx+e\)

Theo bài ta có : \(P\left(x\right)⋮7\Rightarrow\hept{\begin{cases}P\left(0\right)⋮7\\P\left(1\right)⋮7\\P\left(-1\right)⋮7\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}e⋮7\\a+b+c+d+e⋮7\\a-b+c-d+e⋮7\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}a+b+c+d⋮7\\a-b+c-d⋮7\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}a+c⋮7\\b+d⋮7\end{cases}}\)

Mặt khác ta có : \(P\left(2\right)=16a+8b+4c+d+e⋮7\)

\(\Leftrightarrow2a+b+4c+d⋮7\)

\(\Leftrightarrow2\left(a+c\right)+b+d+2c⋮7\)

\(\Leftrightarrow2c⋮7\Leftrightarrow c⋮7\Leftrightarrow a⋮7\)

Chứng minh tương tự thì ta có \(a,b,c,d,e⋮7\). Ta có đpcm.

a)6,6
b)3,5

a) \(\sqrt{40}.\sqrt{12,1}.\sqrt{0,09}=\sqrt{40.12,1.0,09}=\sqrt{\frac{1089}{25}}=\frac{33}{5}\)

b) \(\sqrt{3,5}.\sqrt{2,5}.\sqrt{7}.\sqrt{\frac{1}{5}}=\sqrt{3,5.2,5.7.\frac{1}{5}}=\sqrt{\frac{49}{4}}=\frac{7}{2}\)

a) \(\sqrt{6,4.361}=\sqrt{6,4}.\sqrt{361}=\sqrt{16.0,4}.19\)

\(=\sqrt{16}.\sqrt{0,4}.19=4.\sqrt{0,4}.19=76.\sqrt{0,4}\)

b) \(\sqrt{9,9.1,1}=\sqrt{9.1,1.1,1}=\sqrt{9.1,1^2}=\sqrt{9}.\sqrt{1,1^2}=3.1,1=3,3\)

\(a,\sqrt{6,4.361}=\sqrt{2310,4}=\frac{76\sqrt{10}}{5}\)

\(b,\sqrt{9,9.1,1}=\sqrt{10,89}=3,3\)

Bài làm:

Ta có: \(\left(x-y\right)^2+4\left(x-y\right)-12\)

\(=\left[\left(x-y\right)^2-4\left(x-y\right)+4\right]-16\)

\(=\left(x-y-2\right)^2-4^2\)

\(=\left(x-y-2-4\right)\left(x-y-2+4\right)\)

\(=\left(x-y-6\right)\left(x-y+2\right)\)

\(\left(x-y\right)^2+4\left(x-y\right)-12\)

\(=\left(x-y\right)^2+2.2.\left(x-y\right)+4-16\)

\(=\left(x-y+2\right)^2-4^2\)

\(=\left(x-y+6\right)\left(x-y-2\right)\)

a, \(4x^2-4x-15=\left(2x+3\right)\left(2x-5\right)\)

b, \(2x^2+9x-5=\left(x+5\right)\left(2x-1\right)\)

c, \(3x^2-10x-8=\left(3x+2\right)\left(x-4\right)\)

bạn giải chi tiết hơn 1 chút dc ko bạn

Bài làm:

Ta có: \(x^3+y^3+z^3-3xyz\)

\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)

\(=\left[\left(x+y\right)^3+z^3\right]-\left[3xy\left(x+y\right)+3xyz\right]\)

\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+2xy+y^2-zx-yz+z^2-3xy\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)

và

\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\)

\(=x^2-2xy+y^2+y^2-2yz+z^2+z^2-2zx+x^2\)

\(=2\left(x^2+y^2+z^2-xy-yz-zx\right)\)

Từ đó thay vào P rút ra:

\(P=\frac{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)}{2\left(x^2+y^2+z^2-xy-yz-zx\right)}=\frac{2020}{2}=1010\)

Vậy P = 1010

Gấp lắm ah @Như Trương Thị

P = \(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)

P = \(\frac{2\sqrt{x}-9-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

P = \(\frac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

P = \(\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

P = \(\frac{x-2\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

P = \(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

P = \(\frac{\sqrt{x}+1}{\sqrt{x}-3}\)

Với \(x=6-2\sqrt{5}=5-2\sqrt{5}+1=\left(\sqrt{5}-1\right)^2\) 

=> P = \(\frac{\sqrt{\left(\sqrt{5}-1\right)^2}+1}{\sqrt{\left(\sqrt{5}-1\right)^2}-3}=\frac{\sqrt{5}-1+1}{\sqrt{5}-1-3}=\frac{\sqrt{5}}{\sqrt{5}-4}=\frac{\sqrt{5}\left(\sqrt{5}+4\right)}{\left(\sqrt{5}-4\right)\left(\sqrt{5}+4\right)}=\frac{5+4\sqrt{5}}{-11}\)