75 nhé bạn

TL :

   ( 15 x 15 ) : 2 x ( 15 x 15 ) : 2 x ( 15 x 15 ) : 2 + ( 0 + 1 + 2 + 2 + 3 + 45 ) x 0

= 0

=> vì : tất cả số nào nhân vs 0 đều = 0

_HT_

1423828.125 

        - ht -

a

\(x^2\left(2x+15\right)+4\left(2x+15\right)=0\\ \Leftrightarrow\left(2x+15\right)\left(x^2+4\right)=0\\ \Leftrightarrow2x+15=0\left(x^2+4>0\forall x\right)\\ \Leftrightarrow2x=-15\\ \Leftrightarrow x=-\dfrac{15}{2}\)

b

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

c

\(2\left(x+3\right)-x^2-3x=0\\ \Leftrightarrow2\left(x+3\right)-\left(x^2+3x\right)=0\\ \Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(2-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\2-x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0-3=-3\\x=2-0=2\end{matrix}\right.\)

a: =>(2x+15)(x^2+4)=0

=>2x+15=0

=>2x=-15

=>x=-15/2

b; =>(x-2)(5x-3)=0

=>x=2 hoặc x=3/5

c: =>(x+3)(2-x)=0

=>x=2 hoặc x=-3

`@` `\text {Ans}`

`\downarrow`

`a)`

`x + 10 = 20`

`=> x = 20 -10`

`=> x = 10`

Vậy, `x = 10`

`b)`

`2 * x + 15 = 35`

`=> 2x = 35 - 15`

`=> 2x = 20`

`=> x = 20 \div 2`

`=> x = 10`

Vậy, `x = 10`

`c)`

`3 * ( x + 2 ) = 15`

`=> x + 2 = 15 \div 3`

`=> x + 2 = 5`

`=> x = 5 - 2`

`=> x = 3`

Vậy, `x = 3`

`d)`

`10 * x + 15 * 11 = 20 * 10`

`=> 10x + 165 = 200`

`=> 10x = 200 - 165`

`=> 10x = 35`

`=> x = 35 \div 10`

`=> x = 3,5`

Vậy,` x = 3,5`

`e)`

`4 * ( x + 2 ) = 3 * 4`

`=> x + 2 = 12 \div 4`

`=> x + 2 = 3`

`=> x = 3 - 2`

`=> x = 1`

Vậy,` x = 1`

`f)`

`33 x + 135 = 26 * 9`

`=> 33x + 135 = 234`

`=> 33x = 234 - 135`

`=> 33x = 99`

`=> x = 99 \div 33`

`=> x = 3`

Vậy, `x = 3`

`g)`

`2 * x + 15 + 16 + 17 = 100`

`=> 2x + 48 = 100`

`=> 2x = 100 - 48`

`=> 2x = 52`

`=> x = 52 \div 2`

`=> x =26`

`h)`

`2 * (x + 9 + 10 + 11) = 4 . 12 . 25`

`=> 2 * (x + 9 + 10 + 11) = 4*25*12`

`=> 2 * (x + 9 + 10 + 11) = 100*12`

`=> x + 9 + 10 + 11 = 100*12 \div 2`

`=> x + 30 = 600`

`=> x = 600 - 30`

`=> x = 570`

Vậy, `x = 570.`

a) \(x+10=20\Leftrightarrow x=10\)

b) \(2x+15=35\Leftrightarrow2x=20\Leftrightarrow x=10\)

c) \(3.\left(x+2\right)=15\Leftrightarrow x+2=5\Leftrightarrow x=3\)

d) \(10x+15.11=20.10\Leftrightarrow10x+165=200\Leftrightarrow10x=35\Leftrightarrow x=\dfrac{35}{10}=\dfrac{7}{2}\)

e) \(4.\left(x+2\right)=3.4\Leftrightarrow x+2=3\Leftrightarrow x=1\)

f) \(35x+135=26.9\Leftrightarrow35x=234-135\Leftrightarrow35x=99\Leftrightarrow x=\dfrac{99}{35}\)

g) \(2x+15+16+17=100\Leftrightarrow2x+48=100\Leftrightarrow2x=52\Leftrightarrow x=26\)

h) \(2.\left(x+9+10+11\right)=4.12.25\)

\(\Leftrightarrow x+30=2.12.25\)

\(\Leftrightarrow x=600-30\)

\(\Leftrightarrow x=570\)

a: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow2x=-7\)

hay \(x=-\dfrac{7}{2}\)

b: Ta có: \(\left(x-2\right)^3-\left(x-4\right)\left(x^2+4x+16\right)+6\left(x+1\right)^2=49\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6\left(x+1\right)^2=49\)

\(\Leftrightarrow-6x^2+12x+56+6x^2+12x+6=49\)

\(\Leftrightarrow24x=-13\)

hay \(x=-\dfrac{13}{24}\)

Áp dụng BĐT phụ \(a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Leftrightarrow\left(a-b\right)^2\ge0\)

\(A\ge\dfrac{1}{2}\left(x+y+\dfrac{1}{x}+\dfrac{1}{y}\right)^2\ge\dfrac{1}{2}\left(x+y+\dfrac{4}{x+y}\right)^2=\dfrac{1}{2}\left(1+\dfrac{4}{1}\right)^2=\dfrac{25}{2}\)

Dấu "=" \(x=y=\dfrac{1}{2}\)

c.

ĐKXĐ: \(\left[{}\begin{matrix}x\le-5\\x\ge6\end{matrix}\right.\)

\(\sqrt{\left(x-3\right)\left(x-5\right)}+\sqrt{\left(x-3\right)\left(x+5\right)}=\sqrt{\left(x-3\right)\left(x-6\right)}\)

- Với \(x\ge6\) , do \(x-3>0\) pt trở thành:

\(\sqrt{x-5}+\sqrt{x+5}=\sqrt{x-6}\)

Do \(\left\{{}\begin{matrix}\sqrt{x-5}>\sqrt{x-6}\\\sqrt{x+5}>0\end{matrix}\right.\) \(\Rightarrow\sqrt{x-5}+\sqrt{x+5}>\sqrt{x-6}\) pt vô nghiệm

- Với \(x\le-5\) pt tương đương:

\(\sqrt{\left(3-x\right)\left(5-x\right)}+\sqrt{\left(3-x\right)\left(-x-5\right)}=\sqrt{\left(3-x\right)\left(6-x\right)}\)

Do \(3-x>0\) pt trở thành:

\(\sqrt{5-x}+\sqrt{-x-5}=\sqrt{6-x}\)

\(\Leftrightarrow-2x+2\sqrt{x^2-25}=6-x\)

\(\Leftrightarrow2\sqrt{x^2-25}=x+6\) (\(x\ge-6\))

\(\Leftrightarrow4\left(x^2-25\right)=x^2+12x+36\)

\(\Leftrightarrow3x^2-12x-136=0\Rightarrow x=\dfrac{6-2\sqrt{111}}{3}\)

a.

Kiểm tra lại đề, pt này không giải được

b.

ĐKXĐ: \(x\ge0\)

\(\sqrt{x\left(x+1\right)}-\sqrt{x}+1-\sqrt{x+1}=0\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x+1}-1\right)-\left(\sqrt{x+1}-1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x+1}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x+1}=1\end{matrix}\right.\)

\(\Leftrightarrow...\)

a) x + 10 = 20

<=> x = 20 - 10 = 10

Vậy x = 10

b) 2x + 15 = 35

<=> 2x = 35 - 15 = 20

<=> x = 10

Vậy x = 10

c) 3(x + 2) = 15

<=> x + 2 = 15 : 3 = 5

<=> x = 5 - 2 = 3

Vậy x = 3

d) 10x + 15.11 = 20.10

<=> 10x + 165 = 200

<=> 10x = 200 - 165 = 35

<=> x = 35 : 10 = 3,5

Vậy x = 3,5

e) 4(x + 2) = 3.4

<=> x + 2 = 3

<=> x = 3 - 2 = 1

Vậy x = 1

f) 33x + 135 = 26.9

<=> 33x + 135 = 234

<=> 33x = 234 - 135 = 99

<=> x = 99 : 33 = 3

Vậy x = 3

g) 2x + 15 + 16 + 17 = 100

<=> 2x + 48 = 100

<=> 2x = 100 - 48 = 52

<=> x = 52 : 2 = 26

Vậy x = 26

h) 2(x + 9 + 10 + 11) = 4.12.5

<=> x + 30 = 120

<=> x = 120 - 30 = 90

Vậy x = 90

bnhhjnh

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chịu thôi 2563

Áp dụng bất đẳng thức AM - GM:

\(\sqrt{\left(x^2-15\right)\left(x-3\right)}\le\dfrac{x^2-15+x-3}{2}=\dfrac{x^2+x-18}{2};\sqrt{x^2-15}\le\dfrac{x^2-15+1}{2}=\dfrac{x^2-14}{2};\sqrt{x-3}\le\dfrac{x-3+1}{2}=\dfrac{x-2}{2}\).

Do đó \(F\ge x^2+x-\dfrac{x^2+x-18}{2}-\dfrac{x^2-14}{2}-\dfrac{x-2}{2}-38=-21\).

Đẳng thức xảy ra khi x = 4.

Vậy...