bài tập tết nâng cao phải ko

mk cũng có nhưng chưa làm dc

a/ ĐKXĐ: \(-2\le x\le5\)

\(\sqrt{x+2}+\sqrt{5-x}+\sqrt{\left(x+2\right)\left(5-x\right)}-4=0\)

Đặt \(\sqrt{x+2}+\sqrt{5-x}=a>0\Rightarrow\sqrt{\left(x+2\right)\left(5-x\right)}=\frac{a^2-7}{2}\)

\(\Rightarrow a+\frac{a^2-7}{2}-4=0\)

\(\Leftrightarrow a^2+2a-15=0\Rightarrow\left[{}\begin{matrix}a=3\\a=-5\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{\left(x+2\right)\left(5-x\right)}=\frac{a^2-7}{2}=1\)

\(\Leftrightarrow-x^2+3x+10=1\)

\(\Leftrightarrow x^2-3x-9=0\)

b/ \(\Leftrightarrow\sqrt{x+1}-\sqrt{4-x}+2\left(5+2\sqrt{\left(x+1\right)\left(4-x\right)}\right)=17\)

Đặt \(\sqrt{x+1}-\sqrt{4-x}=a\Rightarrow\sqrt{\left(x+1\right)\left(4-x\right)}=\frac{5-a^2}{2}\)

\(a+2\left(5+5-a^2\right)=17\)

\(\Leftrightarrow-2a^2+a+3=0\Rightarrow\left[{}\begin{matrix}a=-1\\a=\frac{3}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x+1}-\sqrt{4-x}=-1\\\sqrt{x+1}-\sqrt{4-x}=\frac{3}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}+1=\sqrt{4-x}\\2\sqrt{x+1}=2\sqrt{4-x}+3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2+2\sqrt{x+1}=4-x\\4x+4=25-4x+12\sqrt{4-x}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=1-x\left(x\le1\right)\\12\sqrt{4-x}=8x-21\left(x\ge\frac{21}{8}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=\left(1-x\right)^2\\144\left(4-x\right)=\left(8x-21\right)^2\end{matrix}\right.\)

c/ ĐKXĐ: \(0\le x\le1\)

Đặt \(\sqrt{x}+\sqrt{1-x}=a>0\Rightarrow\sqrt{x-x^2}=\frac{a^2-1}{2}\)

\(a^2-1=3\left(a-1\right)\Leftrightarrow a^2-3a+2=0\Rightarrow\left[{}\begin{matrix}a=1\\a=2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x-x^2}=\frac{a^2-1}{2}=0\\\sqrt{x-x^2}=\frac{a^2-1}{2}=\frac{3}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-x^2=0\\x-x^2=\frac{9}{4}\left(vn\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

d/ ĐKXĐ: ...

Đặt \(\left\{{}\begin{matrix}\sqrt{5+2x}=a\ge0\\\sqrt{5-2x}=b\ge0\end{matrix}\right.\) ta được:

\(\left\{{}\begin{matrix}\left(3a-1\right)\left(3b-1\right)=16\\a^2+b^2=10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3ab-\left(a+b\right)=5\\\left(a+b\right)^2-2ab=10\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=3ab-5\\\left(a+b\right)^2-2ab=10\end{matrix}\right.\)

\(\Rightarrow\left(3ab-5\right)^2-2ab=10\)

\(\Leftrightarrow9\left(ab\right)^2-32ab+15=0\Rightarrow\left[{}\begin{matrix}ab=3\\ab=\frac{5}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left(ab\right)^2=9\\\left(ab\right)^2=\frac{25}{81}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}25-4x^2=9\\25-4x^2=\frac{25}{81}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2=4\\x^2=\frac{500}{81}\end{matrix}\right.\)

Yêu cầu đề là gì thế bạn? Bạn cần viết rõ ra thì mọi người mới giúp được chứ????

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

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

\(\Leftrightarrow8x+1=0\Leftrightarrow x=-\frac{1}{8}\)

b, \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)

\(\Leftrightarrow2x+255=0\Leftrightarrow x=-\frac{225}{2}\)

c, \(\left(x+2\right)\left(x-2\right)-x^3-2x=15\)

\(\Leftrightarrow x^2-4-x^3-2x=15\)( vô nghiệm )

d, \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)

\(\Leftrightarrow x^3+9x^2+27x+27-9x^3+6x^2-x+8x^3+1=28\)

\(\Leftrightarrow15x^2+26=0\Leftrightarrow x^2\ne-\frac{26}{15}\)( vô nghiệm )

Tính nhẩm hết á, sai bỏ quá nhá, sắp đi hc ... nên chất lượng hơi kém xíu ~~~ 

a) ( x + 4 )² - ( x + 1 )( x - 1 ) = 16

<=> x² + 8x + 16 - ( x² - 1 ) = 16

<=> x² + 8x + 16 - x² + 1 = 16

<=> 8x + 17 = 16

<=> 8x = -1

<=> x = -1/8

b) ( 2x - 1 )² + ( x + 3 )² - 5( x + 7 )( x - 7 ) = 0

<=> 4x² - 4x + 1 + x² + 6x + 9 - 5( x² - 49 ) = 0

<=> 5x² + 2x + 10 - 5x² + 245 = 0

<=> 2x + 255 = 0

<=> 2x = -255

<=> 2x = -255/2

c) ( x + 2 )( x² - 2x + 4 ) - x( x² + 2 ) = 15

<=> x³ + 2³ - x³ - 2x = 15

<=> 8 - 2x = 15

<=> 2x = -7

<=> x = -7/2

d) ( x + 3 )³ - x( 3x + 1 )² + ( 2x + 1 )( 4x² - 2x + 1 ) = 28

<=> x³ + 9x² + 27x + 27 - x( 9x² + 6x + 1 ) + [ ( 2x )³ + 1³ ] = 28

<=> x³ + 9x² + 27x + 27 - 9x³ - 6x² - x + 8x³ + 1 = 28

<=> 3x² + 26x + 28 = 28

<=> 3x² + 26x = 0

<=> x( 3x + 26 ) = 0

<=> \(\orbr{\begin{cases}x=0\\3x+26=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{26}{3}\end{cases}}\)

`a)4/5+5 1/2 xx (4,5-2)+7/10`

`=4/5+11/2*2,5+7/10`

`=0,8+2,2+0,7`

`=3+0,7=3,7`

`b)125%xx 17/4:(1 5/16-0,5)+2008`

`=1,25xx4,25:13/16+2008`

`=85/13+2008`

`=2014 7/13`

`c)5/11+(16/11+1)`

`=5/11+1+5/11+1`

`=2+10/11=32/11`

`d)3/17+11/4+5/8+14/17+3/8`

`=3/17+14/17+5/8+3/8+11/4`

`=1+1+11/4`

`=19/4`

a) 

\(\dfrac{4}{5}+5\dfrac{1}{2}x\left(4,5-2\right)=\dfrac{7}{10}\)

<=> \(\dfrac{11}{2}x\times2,5=\dfrac{7}{10}-\dfrac{4}{5}=\dfrac{-1}{10}\)

<=> \(\dfrac{55}{4}x=\dfrac{-1}{10}< =>x=\dfrac{-2}{275}\)

b) \(125\%\times\dfrac{17}{4}:\left(1\dfrac{5}{16}-0,5\right)+2008\)

= \(\dfrac{85}{16}:\left(\dfrac{21}{16}-\dfrac{1}{2}\right)+2008=\dfrac{85}{16}:\dfrac{13}{16}+2008=\dfrac{26189}{13}\)

c) \(\dfrac{5}{11}+\left(\dfrac{16}{11}+1\right)\)

= \(\dfrac{21}{11}+1=\dfrac{32}{11}\)

d) \(\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)+\dfrac{11}{4}\)

= 1 + 1 + \(\dfrac{11}{4}\) = \(\dfrac{19}{4}\)

Bài 17)

(x - 2)^4 + (x - 6)^4 = 82
Đặt t = x + 3
=> x + 2 = t - 1; x + 4 = t + 1.
ta có pt: (t - 1)^4 + (t + 1)^4 = 82
<=>[(t -1)²]² + [(t + 1)²]² = 82
<=> (t² - 2t + 1)² + (t² + 2t + 1)² = 82
<=> (t²+1)² - 4t(t²+1) + 4t² + (t²+1)² + 4t(t²+1) + 4t² = 82
<=> (t² + 1)² + 4t² = 41
<=> t^4 + 6t² + 1 = 41
<=> (t²)² + 6t² - 40 = 0
<=> t² = -10 (loại) hoặc t² = 4
<=> t = 2 hoặc t = -2
với t = -2 => x = -5
với t = 2 => x = -1
vậy pt có hai nghiệm là : x = -1 hoặc x = -5

Bài 18: Phương trình đã cho được viết thành: $${({x^2} + 6x + 10)^2} + (x + 3)\left[ {3\left( {{x^2} + 6x + 10} \right) + 2\left( {x + 3} \right)} \right] = 0$$
Đặt $u = {x^2} + 6x + 10 > 0,v = x + 3$, suy ra:
$${u^2} + v\left( {3u + 2v} \right) = 0 \Leftrightarrow \left( {u + v} \right)\left( {u + 2v} \right) = 0 \Leftrightarrow \left[ \begin{gathered}
u + v = 0 \\
u + 2v = 0 \\
\end{gathered} \right.$$
$$ \Leftrightarrow \left[ \begin{gathered}
{x^2} + 6x + 10 + x + 3 = 0 \\
{x^2} + 6x + 10 + 2\left( {x + 3} \right) = 0 \\
\end{gathered} \right. \Leftrightarrow \left[ \begin{gathered}
{x^2} + 7x + 13 = 0 \\
{x^2} + 8x + 16 = 0 \\
\end{gathered} \right. \Leftrightarrow x = - 4$$

Bài 19:

(x² + x + 1) - 7(x - 1)² = 13(x³ - 1)
⇔ 2x² + 2x + 2 - 7(x² - 2x + 1) = 13x - 13
⇔ 2x² + 2x + 2 - 7x² + 14x - 7 = 13x³ - 13
⇔ 13x³ + 5x² - 16x - 8 = 0
⇔ 13x³ + 13x² - 8x² - 8x - 8x - 8 = 0
⇔ 13x²(x + 1) - 8x(x + 1) - 8(x + 1) = 0
⇔ (x + 1)(13x² - 8x - 8) = 0
⇔ \(\left[{}\begin{matrix}x+1=0\\13x^2-8x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{4\pm2\sqrt{30}}{13}\end{matrix}\right.\)

len google di ban

mk chua hoc bai nay

d) \(\left(x+2\right)\left(x^2-2x+4\right)\)

\(=\left(x+2\right)\left(x^2-2\cdot x+2^2\right)\)

\(=x^3+2^3\)

\(=x^3+8\)

e) \(\left(\dfrac{1}{4}-\dfrac{x}{5}\right)\left(\dfrac{x^2}{25}+\dfrac{x}{20}+\dfrac{1}{16}\right)\)

\(=\left(\dfrac{1}{4}-\dfrac{1}{5}x\right)\left(\dfrac{1}{25}x^2+\dfrac{1}{5}x\cdot\dfrac{1}{4}+\dfrac{1}{16}\right)\)

\(=\left(\dfrac{1}{4}-\dfrac{1}{5}x\right)\left[\left(\dfrac{1}{5}x\right)^2+\dfrac{1}{5}x\cdot\dfrac{1}{4}+\left(\dfrac{1}{4}\right)^2\right]\)

\(=\left(\dfrac{1}{4}\right)^3-\left(\dfrac{1}{5}x\right)^3\)

\(=\dfrac{1}{64}-\dfrac{1}{125}x^3\)

\(=\dfrac{1}{64}-\dfrac{x^3}{125}\)

d: (x+2)(x^2-2x+4)

=(x+2)(x^2-x*2+2^2)

=x^3+8

e: (1/4-x/5)(1/16+x/20+x^2/25)

=(1/4-x/5)[(1/4)^2+1/4*x/5+(x/5)^2]

=1/64-x^3/125