a, 7x + 10x  = 5x 

    17x = 5x

17x - 5x = 0

      12x = 0

          x =0

2; 

a, 4x + 7x = 22

    11x = 22

        x = 2

b, 12x - 8x = 25

     4x = 25

       x = \(\dfrac{25}{4}\)

c,  \(\dfrac{1}{2}\)x - \(\dfrac{1}{3}\)x = \(\dfrac{4}{5}\) 

     (\(\dfrac{1}{2}-\dfrac{1}{3}\))x = \(\dfrac{4}{5}\)

    \(\dfrac{1}{6}\)x     = \(\dfrac{4}{5}\) 

      x = \(\dfrac{4}{5}\) : \(\dfrac{1}{6}\)

     x = \(\dfrac{24}{5}\)

ĐKXĐ: \(x\ge\dfrac{2}{3}\)

\(\Leftrightarrow x\sqrt{3x-2}-x^2+\left(x+1\right)\sqrt{5x-1}-\left(x+1\right)^2+x^2+\left(x+1\right)^2-8x+3=0\)

\(\Leftrightarrow x\left(\sqrt{3x-2}-x\right)+\left(x+1\right)\left(\sqrt{5x-1}-x-1\right)+2\left(x^2-3x+2\right)=0\)

\(\Leftrightarrow\dfrac{-x\left(x^2-3x+2\right)}{\sqrt{3x-2}+x}+\dfrac{-\left(x+1\right)\left(x^2-3x+2\right)}{\sqrt{5x-1}+x+1}+2\left(x^2-3x+2\right)=0\)

\(\Leftrightarrow\left(x^2-3x+2\right)+\left(2-\dfrac{x}{\sqrt{3x-2}+x}-\dfrac{x+1}{\sqrt{5x-1}+x+1}\right)=0\)

\(\Leftrightarrow\left(x^2-3x+2\right)\left(\dfrac{\sqrt{3x-2}}{\sqrt{3x-2}+x}+\dfrac{\sqrt{5x-1}}{\sqrt{5x-1}+x+1}\right)=0\)

\(\Leftrightarrow x^2-3x+2=0\) (ngoặc đằng sau luôn dương)

\(\Leftrightarrow...\)

1)

ĐKXĐ: x\(\ne\)3

ta có :

\(\frac{x^2-6x+9}{2x-6}=\frac{\left(x-3\right)^2}{2\left(x-3\right)}=\frac{x-3}{2}\)

để biểu thức A có giá trị = 1

thì :\(\frac{x-3}{2}\)=1

=>x-3 =2

=>x=5(thoả mãn điều kiện xác định)

vậy để biểu thức A có giá trị = 1 thì x=5

1)

\(A=\frac{x^2-6x+9}{2x-6}\)

A xác định

\(\Leftrightarrow2x-6\ne0\)

\(\Leftrightarrow2x\ne6\)

\(\Leftrightarrow x\ne3\)

Để A = 1

\(\Leftrightarrow x^2-6x+9=2x-6\)

\(\Leftrightarrow x^2-6x-2x=-6-9\)

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

\(\Leftrightarrow x=3\) (loại vì không thỏa mãn ĐKXĐ)

Có : \(2x^2+3x+2\)

\(\Leftrightarrow\) \(\left(x^2+2x+1^2\right)+\left(x^2+x+1^2\right)\)

\(\Leftrightarrow\) \(\left(x^2+2.x.1+1^2\right)\) + \(\left(x^2+2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2+\frac{3}{4}\right)\)

\(\Leftrightarrow\) \(\left(x+1\right)^2+\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)

Vì \(\left(x+1\right)^2\ge0và\left(x+\frac{1}{2}\right)^2\ge0\)

\(\Rightarrow\) \(\left(x+1\right)^2+\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)

Vậy \(2x^2+3x+2>0\left(\forall_x\right)\)

a)

\(2x^2+3x+2=\left(x^2+2x+1\right)+\left(x^2+2\cdot\frac{1}{2}x+\frac{1}{4}\right)+\frac{3}{4}\\ =\left(x+1\right)^2+\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)

(vì >3/4 nên >0)

`@` `\text {Ans}`

`\downarrow`

`1)`

\(2x+\dfrac{1}{2}=\dfrac{5}{3}\)

`\Rightarrow`\(2x=\dfrac{5}{3}-\dfrac{1}{2}\)

`\Rightarrow`\(2x=\dfrac{7}{6}\)

`\Rightarrow`\(x=\dfrac{7}{6}\div2\)

`\Rightarrow`\(x=\dfrac{7}{12}\)

Vậy, `x = 7/12`

`2)`

\(\dfrac{1}{7}+\dfrac{4}{5}x=\dfrac{5}{3}\)

`\Rightarrow`\(\dfrac{4}{5}x=\dfrac{5}{3}-\dfrac{1}{7}\)

`\Rightarrow`\(\dfrac{4}{5}x=\dfrac{32}{21}\)

`\Rightarrow`\(x=\dfrac{32}{21}\div\dfrac{4}{5}\)

`\Rightarrow`\(x=\dfrac{40}{21}\)

Vậy, `x = 40/21`

`3)`

\(\dfrac{3}{5}-\dfrac{3}{5}x=\dfrac{1}{7}\)

`\Rightarrow`\(\dfrac{3}{5}x=\dfrac{3}{5}-\dfrac{1}{7}\)

`\Rightarrow`\(\dfrac{3}{5}x=\dfrac{16}{35}\)

`\Rightarrow`\(x=\dfrac{16}{35}\div\dfrac{3}{5}\)

`\Rightarrow`\(x=\dfrac{16}{21}\)

Vậy, `x = 16/21`

`4)`

\(\dfrac{5}{6}-3x=\dfrac{3}{4}\)

`\Rightarrow`\(3x=\dfrac{5}{6}-\dfrac{3}{4}\)

`\Rightarrow`\(3x=\dfrac{1}{12}\)

`\Rightarrow`\(x=\dfrac{1}{12}\div3\)

`\Rightarrow`\(x=\dfrac{1}{36}\)

Vậy, `x  = 1/36`

`5)`

\(\dfrac{5}{3}-\dfrac{1}{2}x=\dfrac{3}{7}\)

`\Rightarrow`\(\dfrac{1}{2}x=\dfrac{5}{3}-\dfrac{3}{7}\)

`\Rightarrow`\(\dfrac{1}{2}x=\dfrac{26}{21}\)

`\Rightarrow`\(x=\dfrac{26}{21}\div\dfrac{1}{2}\)

`\Rightarrow`\(x=\dfrac{52}{21}\)

Vậy, `x = 52/21`

`6)`

\(5x+\dfrac{1}{2}=\dfrac{2}{3}\)

`\Rightarrow`\(5x=\dfrac{2}{3}-\dfrac{1}{2}\)

`\Rightarrow`\(5x=\dfrac{1}{6}\)

`\Rightarrow`\(x=\dfrac{1}{6}\div5\)

`\Rightarrow`\(x=\dfrac{1}{30}\)

Vậy, `x = 1/30.`

a) ( x - 3)⁴ + ( x - 5)⁴ = 82

Đặt : x - 4 = a , ta có :

( a + 1)⁴ + ( a - 1)⁴ = 82

⇔ a⁴ + 4a³ + 6a² + 4a + 1 + a⁴ - 4a³ + 6a² - 4a + 1 = 82

⇔ 2a⁴ + 12a² - 80 = 0

⇔ 2( a⁴ + 6a² - 40) = 0

⇔ a⁴ - 4a² + 10a² - 40 = 0

⇔ a²( a² - 4) + 10( a² - 4) = 0

⇔ ( a² - 4)( a² + 10) = 0

Do : a² + 10 > 0

⇒ a² - 4 = 0

⇔ a = + - 2

+) Với : a = 2 , ta có :

x - 4 = 2

⇔ x = 6

+) Với : a = -2 , ta có :

x - 4 = -2

⇔ x = 2

KL.....

b) ( n - 6)( n - 5)( n - 4)( n - 3) = 5.6.7.8

⇔ ( n - 6)( n - 3)( n - 5)( n - 4) = 1680

⇔ ( n² - 9n + 18)( n² - 9n + 20) = 1680

Đặt : n² - 9n + 19 = t , ta có :

( t - 1)( t + 1) = 1680

⇔ t² - 1 = 1680

⇔ t² - 41² = 0

⇔ ( t - 41)( t + 41) = 0

⇔ t = 41 hoặc t = - 41

+) Với : t = 41 , ta có :

n² - 9n + 19 = 41

⇔ n² - 9n - 22 = 0

⇔ n² + 2n - 11n - 22 = 0

⇔ n( n + 2) - 11( n + 2) = 0

⇔ ( n + 2)( n - 11) = 0

⇔ n = - 2 hoặc n = 11

+) Với : t = -41 ( giải tương tự )

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