(x-7)(x-5)(x-4)(x-2)=72

<=> (x-7)(x-2)(x-5)(x-4)=72

<=> (x^2-9x+14)(x^2-9x+20)=72

đặt t=x^2-9x+17 (1)

pt trở thành

(t-3)(t+3)=72

<=> t^2-81=0<=> t^2=81<=> t=9 hoặc t=-9

thế t vào (1)

th1 x^2-9x+17=9

<=> x^2-9x+8=0

giải pt => x=8 hoặc x=1

th2 x^2-9x+17=-9

<=> x^2-9x+26=0

giải pt => pt vô nghiệm 

S={8;1}

\(\left(x-7\right)\left(x-2\right)\left(x-5\right)\left(x-4\right)\)
\(\left(x^2-9x+14\right)\left(x^2-9x+20\right)=72\)

Đặt : \(x^2-9x+14=t\left(t>0\right)\)

\(\Rightarrow t\left(t+6\right)-72=0\Rightarrow t_1=6\left(tm\right)'t_2=-12\)(loại)

Với \(t=6\Rightarrow x^2-9x+14=6\)

\(\Rightarrow x_1=8;x_2=1\)

ô ze để tui giải chờ tí nhé

Ta có (x-7)(x-2)(x-5)(x-4)=72 
<=> (x²-9x+14)(x²-9x+20)=72 
Đặt x²-9x+14 = t (đk t>0) 
=> t(t+6) - 72 = 0

=> t₁=6 (thỏa mãn)  và t₂ = -12 (loại) 
Khi t=6 => x²-9x+14 = 6 
=> x₁ = 8 ; x₂ = 1

\(\left(x-7\right)\left(x-2\right)\left(x-5\right)\left(x-4\right)\)

\(\left(x^2-9x+14\right)\left(x^2-9x+20\right)=72\)

Đặt : \(x^2-9x+14=t\left(t>0\right)\)

\(\Rightarrow t\left(t+6\right)-72=0\Rightarrow t_1=6\left(tm\right)'t_2=-12\left(loại\right)\)

Với : \(t=6\Rightarrow x^2-9x+14=6\)

\(\Rightarrow x_1=8;x_2=1\)

(x-7)(x-5)(x-4)(x-2)=72

<=>[(x-7)(x-2)].[(x-5)(x-4)]=72

<=>(x²-9x+14)(x²-9x+20)=72

<=>(x²-9x+17-3)(x²-9x+17+3)=72 

Đặt m=x²-9x+17 (*)

PT <=> (m-3)(m+3)=72

<=>m²-3²=72<=>m²-9=72<=>m²=81<=> m E {-9;9}

+Với m=-9 ,thay vào (*),ta được:x²-9x+17=-9

=>x²-9x+17-(-9)=0=>x²-9x+26=0

=>x²-\(2.x.\frac{9}{2}\) + \(\left(\frac{9}{2}\right)^2+\frac{23}{4}\)=0

=> \(\left(x-\frac{9}{2}\right)^2+\frac{23}{4}=0\)

Vì \(\left(x-\frac{9}{2}\right)^2\) > 0 với mọi x

=> \(\left(x-\frac{9}{2}\right)^2+\frac{23}{4}\) > 23/4  > 0

=>PT vô nghiệm

+m=9,thay vào PT (*),ta đc: x²-9x+17=9

=>x²-9x+17-9=0

=>x²-9x+8=0

=>x²-x-8x+8=0

=>x(x-1)-8(x-1)=0

=>(x-1)(x-8)=0

=>x=1 hoặc x=8

Vậy PT có tập nghiệm S = {1;8}

a. (x-1)x(x+1)(x+2)=24

[(x-1)(x+2)].[x(x+1)]=24

(\(x^2\)+2x-x-2)(\(x^2\)+x)=24

(\(x^2\)+x-2)(\(x^2\)+x)=24

[(\(x^2\)+x-1)-1].[(\(x^2\)+x-1)+1]=24

\(\left(x^2+x-1\right)^2\)-1=24

\(\left(x^2+x-1\right)^2\)=25

\(\left(x^2+x-1\right)^2\)=\(5^2\) hoặc\(\left(x^2+x-1\right)^2\)=\(\left(-5\right)^2\)

\(x^2\)+x-1=5 hoặc \(x^2\)+x-1=-5

\(x^2\)+x-6=0 hoặc \(x^2\)+x+4=0(vô nghiệm)

\(\left[\begin{array}{nghiempt}x=2\\x=-3\end{array}\right.\)

Vậy x=2 hoặc x=-3

a)(x-1)x=x²-x

(x+1)(x+2)=x(x+2)+(x+2)=x²+2x+x+2=x²+3x+2

=>(x-1)x(x+1)(x+2)=(x²-x)(x²+3x+2)=x²(x²+3x+2)-x(x²+3x+2)=x⁴+3x³+2x²-x³-3x²-2x

=x⁴+2x³-x²-2x

mà (x-1)x(x+1)(x+2)=24

nên x⁴+2x³-x²-2x=24

x³(x+2)-x(x+2)=24

(x³-x)(x+2)=24

Ta xét bảng sau:

(ô trống là loại)

Vậy x=2, hờ hờ, t làm tầm bậy, không theo phương trình gì hết

ũa,Linh ,c thiếu x=-3 kìa
 

Cách 1:

GPT :\(5\sqrt{x-1}-\sqrt{x+7}=3x-4\) - Hoc24

Cách 2:

Đặt \(\left\{{}\begin{matrix}\sqrt{25x-25}=a\\\sqrt{x+7}=b\end{matrix}\right.\)  \(\Rightarrow3x-4=\dfrac{a^2-b^2}{8}\)

Pt trở thành:

\(a-b=\dfrac{a^2-b^2}{8}\)

\(\Leftrightarrow\left(a-b\right)\left(a+b-8\right)=0\)

\(\Leftrightarrow...\)

\(x^{11}+3x^{10}+x^9+3x^8+x^7-3x^6-17x^5+3x^4+x^3+3x^2+x+3=0\)

\(\Leftrightarrow\left(x^{11}+2x^{10}+4x^9+6x^8+9x^7+6x^6+4x^5+2x^4+x^3\right)+\left(x^{10}+2x^9+4x^8+6x^7+9x^6+6x^5+4x^4+2x^3+x^2\right)-\left(5x^9+10x^8+20x^7+30x^6+45x^5+30x^4+20x^3+10x^2+5x\right)+\left(3x^8+6x^7+12x^6+18x^5+27x^4+18x^3+12x^2+6x+3\right)=0\)

\(\Leftrightarrow x^3\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)+x^2\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)-5\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)+3\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)=0\)

\(\Leftrightarrow\left(x^3+x^2-5x+3\right)\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)=0\)

\(\Leftrightarrow\left(x^2-2x+1\right)\left(x+3\right)\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2\left(x+3\right)\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)=0\)

Dễ thấy: \(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1>0\forall x\)

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

đex ~ vừa thấy trên face lướt qua luôn

7 x 8 = 56       16 : 2 = 8       36 : 6 = 6       49 : 7 = 7

2 x 5 = 10       72 : 8 = 9       9 x 3 = 27       63 : 7 = 9

6 x 4 = 24       25 : 5 = 5       4 x 8 = 32       7 x 5 = 35

7 x 8 = 56

2 x 5 = 10

6 x 4 = 24

16 : 2 = 8 

72 : 8 = 9

25 : 5 = 5

36 : 6 = 6

9 x 3 = 27

4 x 8 = 32

49 : 7 = 7

63 : 7 = 9

7 x 5 = 35

ĐKXĐ: \(x\ge\sqrt[3]{7}\)

\(\sqrt{x^4-7}-\left(x^2-1\right)+\sqrt{x^3-7}-1=0\)

\(\Leftrightarrow\dfrac{x^4-7-\left(x^2-1\right)^2}{\sqrt{x^4-7}+\left(x^2-1\right)}+\dfrac{x^3-8}{\sqrt{x^3-7}+1}=0\)

\(\Leftrightarrow\dfrac{2\left(x^2-4\right)}{\sqrt{x^4-7}+\left(x^2-1\right)}+\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{\sqrt{x^3-7}+1}=0\)

\(\Leftrightarrow\dfrac{2\left(x-2\right)\left(x+2\right)}{\sqrt{x^4-7}+\left(x^2-1\right)}+\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{\sqrt{x^3-7}+1}=0\)

\(\Leftrightarrow\left(x-2\right)\left(\dfrac{2\left(x+2\right)}{\sqrt{x^4-7}+\left(x^2-1\right)}+\dfrac{x^2+2x+4}{\sqrt{x^3-7}+1}\right)=0\)

Do \(x\ge\sqrt[3]{7}>1\Rightarrow x^2>1\Rightarrow x^2-1>0\)

\(\Rightarrow\dfrac{2\left(x+2\right)}{\sqrt{x^4-7}+\left(x^2-1\right)}+\dfrac{x^2+2x+4}{\sqrt{x^3-7}+1}>0\)

\(\Rightarrow x-2=0\Rightarrow x=2\)

Vậy pt có nghiệm duy nhất \(x=2\)

 (x-7)(x-2)(x-5)(x-4)=72 
(x^2-9x+14)(x^2-9x+20)=72 
đặt x^2-9x+14=t (t>0) 
=> t(t+6)-72=0 =>t1=6 (tm) ' t2=-12 (loại) 
với t=6 => x^2-9x+14=6 
=> x1=8 ; x2=1