\(\left(x+4\right)^2-\left(x+1\right)\left(x+1\right)=16.\)

\(\Leftrightarrow x^2+8x+16-x^2-2x-1=16.\)

\(\Leftrightarrow6x+15=16.\Leftrightarrow x=\dfrac{1}{6}.\)

\(3x=2y=7z\)

⇒\(\dfrac{x}{14}=\dfrac{y}{21}=\dfrac{z}{6}\)

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

\(\dfrac{x}{14}=\dfrac{z}{6}=\dfrac{x-z}{14-6}=\dfrac{16}{8}=2\)

⇒\(\left\{{}\begin{matrix}x=14.2=28\\y=21.2=42\\z=6.2=12\end{matrix}\right.\)

G(-2)=0

=>4a+8+16=0

=>4a=-24

hay a=-6

đặt g(-2) = 0

=> a.(-2)² - 4.(-2) + 16 = 0

<=> 4a + 8 + 16 = 0

<=> 4a = -24

<=> a = -6

vậy để x = -2 là nghiệm của g(x) thì a = -6

a: ĐKXĐ: 3x+24<>0

=>3x<>-24

=>x<>-8

b: ĐKXĐ: \(x^2-16<>0\)

=>\(x^2<>16\)

=>x∉{4;-4}

`#3107.101107`

a)

`x^2 + 6x + 10`

`= (x^2 + 2*x*3 + 3^2) + 1`

`= (x + 3)^2 + 1`

Vì `(x + 3)^2 \ge 0` `AA` `x`

`=> (x + 3)^2 + 1 \ge 1` `AA` `x`

Vậy, GTNN của bt là 1 khi `(x + 3)^2 = 0`

`<=> x + 3 = 0`

`<=> x = -3`

b)

`4x^2 - 4x + 5`

`= [(2x)^2 - 2*2x*1 + 1^2] + 4`

`= (2x - 1)^2 + 4`

Vì `(2x - 1)^2 \ge 0` `AA` `x`

`=> (2x - 1)^2 + 4 \ge 4` `AA` `x`

Vậy, GTNN của bt là `4` khi `(2x - 1)^2 = 0`

`<=> 2x - 1 = 0`

`<=> 2x = 1`

`<=> x = 1/2`

c)

`x^2 - 3x + 1`

`= (x^2 - 2*x*3/2 + 9/4) - 5/4`

`= (x - 3/2)^2 - 5/4`

Vì `(x - 3/2)^2 \ge 0` `AA` `x`

`=> (x - 3/2)^2 - 5/4 \ge -5/4` `AA` `x`

Vậy, GTNN của bt là `-5/4` khi `(x - 3/2)^2 = 0`

`<=> x - 3/2 = 0`

`<=> x = 3/2`

=>5x^3+4x^2+3x+3-4+x+4x^2-5x^3=5

=>8x^2+4x-1-5=0

=>8x^2+4x-6=0

=>4x^2+2x-3=0

=>\(x=\dfrac{-1\pm\sqrt{13}}{4}\)

1: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-\left(x^3+54\right)\)

\(=x^3+27-x^3-54\)

=-27

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

\(=8x^3+y^3-8x^3+y^3\)

\(=2y^3\)​

\(1,=x^3+270-x^3-54=-27\\ 2,=8x^3+y^3-8x^3+y^3=2y^3\\ 3,=x^3-3x^2+3x-1-x^3-8+3x^2-48=3x-57\\ 4,=x^3-x-x^3-1=-x-1\\ 5,=8x^3-5\left(8x^3+1\right)=-32x^3-5\\ 6,=27+x^3-27=x^3\\ 7,làm.ở.câu.3\\ 8,=x^3-6x^2+12x-8+6x^2-12x+6-x^3-1+3x\\ =3x-3\)

Bài 1:

a: Cách 1: \(\left(\frac23-\frac45\right)\times25\)

\(=\left(\frac{10}{15}-\frac{12}{25}\right)\times25\)

\(=-\frac{2}{25}\times15=-\frac{30}{15}=-\frac65\)

Cách 2: \(\left(\frac23-\frac45\right)\times25\)

\(=\frac23\times25-\frac45\times25\)

\(=\frac{50}{3}-20=\frac{50}{3}-\frac{100}{5}=\frac{50}{3}-20=\frac{50}{3}-\frac{60}{3}=-\frac{10}{3}\)

b: Sửa đề: \(\frac34\times\frac25+\frac34\times\frac27\)

Cách 1: \(\frac34\times\frac25+\frac34\times\frac27\)

\(=\frac34\times\left(\frac25+\frac27\right)\)

\(=\frac34\times\left(\frac{14}{35}+\frac{10}{35}\right)\)

\(=\frac34\times\frac{24}{35}=\frac{72}{140}=\frac{18}{35}\)

Cách 2:

\(\frac34\times\frac25+\frac34\times\frac27\)

\(=\frac{6}{20}+\frac{6}{28}\)

\(=\frac{42}{140}+\frac{30}{140}=\frac{72}{140}=\frac{18}{35}\)

Bài 2:

a: \(x:\frac25=\frac23\)

=>\(x=\frac23\times\frac25=\frac{4}{15}\)

b: \(x:\frac23=\frac49\)

=>\(x=\frac49\times\frac23=\frac{8}{27}\)