ai giải hộ với aaaa

Đặt x - 7 = t => x - 8 = t - 1, 15 - 2x = -2t + 1 
thay vào pt được: 
----> t^4 + (t - 1)^4 = (-2t + 1)^4 
<=> t^4 + t^4 - 4t³ + 6t² - 4t + 1 = 16t^4 - 16t³ + 24t² - 8t + 1 
<=> 14t^4 - 12t³ + 18t² - 4t = 0 
<=> t( 14t³ - 12t² + 18t - 4) = 0 
<=> t = 0 hoặc 14t³ - 12t² + 18t - 4 = 0 
+ Với t = 0 => x - 7 = 0 <=> x = 7 
+ Với 14t³ - 12t² + 18t - 4 = 0 --->pt vô n_o

<=> x-7=7

<=> x=14 

--> S={14}

đúng tích mk nha

Theo bài ra , ta có : 

\(\left(x-7\right)^4+\left(x-8\right)^4=\left(15-2x\right)^4\)

\(\Leftrightarrow x^4-28x^3+294x^2-1372x+2401+x^4-32x^3+384x^2-2048x+4096=\left(15-2x\right)^4\)

\(\Leftrightarrow2x^4-60x^3+678x^2-3420x+6497=50625-27000x+5400x^2-480x^3+16x^4\)

\(\Leftrightarrow-14x^4+420x^3-4722x^2+23580x=44128\)

\(-2x\left(7x^3-210x^2+2361x-11790\right)=44128\)

\(\Leftrightarrow-2x\left(\left(x-15\right)\left(7x^2-105x+786\right)\right)=44128\)

\(\Leftrightarrow x=8\)

Vậy tập nghiệm của phương trình là \(S=\left\{8\right\}\)

Chúc bạn học tốt =)) 

x={7;8}

Đặt: \(\hept{\begin{cases}x-7=a\\x-8=b\end{cases}\Rightarrow}2x-15=a+b\)

khi đó pt trở thành: \(a^4+b^4=\left(a+b\right)^4\)

\(\Leftrightarrow a^4+b^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4\)

\(\Leftrightarrow4a^3b+6a^2b^2+4ab^3=0\)

\(\Leftrightarrow2ab\left(2a^2+3ab+2b^2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}ab=0\\2a^2+3ab+b^2=0\end{cases}}\)

TH1: \(ab=0\Leftrightarrow\orbr{\begin{cases}a=0\\b=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x-7=0\\x-8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=8\end{cases}}}\)

TH2: \(2a^2+3ab+2b^2=2\left(a^2+\frac{3}{2}ab+b^2\right)=2\left(a^2+2.a.\frac{3}{4}b+\frac{9}{16}b^2+\frac{7}{16}b^2\right)=2\left(a+\frac{3}{4}b\right)^2+\frac{7}{8}b^2\ge0\)Dấu = xảy ra <=> a=b=0 

hay x-7=x-8=0 (vô nghiệm)

Vậy x=7 hoặc x=8 là nghiệm 

\(\left(x-7\right)^4+\left(x-8\right)^4=\left(15-2x\right)^4\)

Đặt \(\hept{\begin{cases}x-7=a\\x-8=b\end{cases}\Rightarrow a+b=2x-15}\)

Ta có:

\(a^4+b^4=\left(a+b\right)^4\)

\(\Leftrightarrow2ab^3+3a^2b^2+2a^3b=0\)

\(\Leftrightarrow ab\left(2a^2+3ab+2b^2\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}a=0\\b=0\\2a^2+3ab+b^2=0\end{cases}}\)

Với \(a=0\Rightarrow x=7\)

Với \(b=0\Rightarrow x=8\)

Với \(2a^2+3ab+b^2=0\) thì ta nhận xét thấy 

\(2a^2+3ab+b^2\ge0\) nhưng dấu = không xảy ra nên phương trình này vô nghiệm.

Vậy ... 

\(\left(x-7\right)^4+\left(x-8\right)^4=\left(15-2x\right)^4\)

\(\Leftrightarrow\left(x-7\right)2^2+\left(x-8\right)2^2=\left(15-2x\right)2^2\)

\(\Leftrightarrow\left(2x-14\right)^2+\left(2x-16\right)^2=\left(30-4x\right)^2\)

\(\Leftrightarrow4x^2-56x+196+4x^2-64x+256=\left(30-4x\right)^2\)

\(\Leftrightarrow8x^2-120x+452=900-240x+16x^2\)

\(\Leftrightarrow8x^2-120x+452-900+240x-16x^2=0\)

\(\Leftrightarrow-8x^2+120x-448=0\)

\(\Leftrightarrow-\left(8x^2-120x+448\right)=0\)

tự làm tiếp nha

Chuyển (15-2x)^4 sang vế bên kia.

Đặt VT=f(x) (VT sau khi đã chuyển (15-2x)^4).

Xét dấu của f(x), nhận 2 nghiệm x=7 ; x=8.

\(a,2x\left(x-5\right)+4\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\2x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\2x=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{5;-2\right\}\)

\(b,3x-15=2x\left(x-5\right)\\ \Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(-2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\-2x+3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{5;\dfrac{3}{2}\right\}\)

\(c,\left(2x+1\right)\left(3x-2\right)=\left(5x-8\right)\left(2x+1\right)\\ \Leftrightarrow\left(2x+1\right)\left(3x-2\right)-\left(5x-8\right)\left(2x+1\right)=0\\ \Leftrightarrow\left(2x+1\right)\left(3x-2-5x+8\right)=0\\ \Leftrightarrow\left(2x+1\right)\left(-2x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+1=0\\-2x+6=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=-1\\2x=6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)

Vậy \(x\in\left\{-\dfrac{1}{2};3\right\}\)

Câu d xem lại đề

a: =>x^2-8x+16<x^2-8x

=>16<0(loại)

b: =>\(x+\dfrac{1}{2}>=\dfrac{5x-3}{3}\)

=>x+1/2>=5/3x-1

=>-2/3x>=-3/2

=>x<=3/2:2/3=9/4

c: =>\(\dfrac{7-x}{4}< =\dfrac{2x-4}{3}\)

=>21-3x<=8x-16

=>-11x<=-37

=>x>=37/11

(x - 7)^4 + (x - 8)^4 = (15 - 2x)^4
Đặt x - 7 = t

\(\Rightarrow\)x - 8 = t - 1 và 15 - 2x = -2t + 1
thay vào pt được:
\(\rightarrow\)t^4 + (t - 1)^4 = (-2t + 1)^4
\(\Leftrightarrow\) t^4 + t^4 - 4t³ + 6t² - 4t + 1 = 16t^4 - 16t³ + 24t² - 8t + 1
\(\Leftrightarrow\) 14t^4 - 12t³ + 18t² - 4t = 0
\(\Leftrightarrow\) t( 14t³ - 12t² + 18t - 4) = 0
\(\Leftrightarrow\) t = 0 hoặc 14t³ - 12t² + 18t - 4 = 0
+) Với t = 0\(\Leftrightarrow\) x - 7 = 0 \(\Leftrightarrow\) x = 7
+ )Với 14t³ - 12t² + 18t - 4 = 0 \(\Rightarrow\) pt vô nghiệm

\(\rightarrow\) S={7}

`1)(x+2)(x+3)(x-7)(x-8)=144`
`<=>[(x+2)(x-7)][(x+3)(x-8)]=144`
`<=>(x^2-5x-14)(x^2-5x-24)=144`
`<=>(x^2-5x-19)^2-25=144`
`<=>(x^2-5x-19)^2-169=0`
`<=>(x^2-5x-6)(x^2-5x-32)=0`
`+)x^2-5x-6=0`
`<=>` $\left[ \begin{array}{l}x=6\\x=-1\end{array} \right.$
`+)x^2-5x-32=0`
`<=>` $\left[ \begin{array}{l}x=\dfrac{5+3\sqrt{17}}{2}\\x=\dfrac{5-3\sqrt{17}}{2}\end{array} \right.$
Vậy `S={-1,6,\frac{5+3\sqrt{17}}{2},\frac{5-3\sqrt{17}}{2}}`

1: Ta có: \(\left(x+2\right)\left(x+3\right)\left(x-7\right)\left(x-8\right)=144\)

\(\Leftrightarrow\left(x^2-7x+2x-14\right)\left(x^2-8x+3x-24\right)=144\)

\(\Leftrightarrow\left(x^2-5x-14\right)\left(x^2-5x-24\right)-144=0\)

\(\Leftrightarrow\left(x^2-5x\right)^2-38\left(x^2-5x\right)+336-144=0\)

\(\Leftrightarrow\left(x^2-5x\right)^2-38\left(x^2-5x\right)+192=0\)

\(\Leftrightarrow\left(x^2-5x\right)^2-6\left(x^2-5x\right)-32\left(x^2-5x\right)+192=0\)

\(\Leftrightarrow\left(x^2-5x\right)\left(x^2-5x-6\right)-32\left(x^2-5x-6\right)=0\)

\(\Leftrightarrow\left(x^2-5x-6\right)\left(x^2-5x-32\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+1\right)\left(x^2-5x-32\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x+1=0\\x^2-5x-32=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-1\\x=\dfrac{5-3\sqrt{17}}{2}\\x=\dfrac{5+3\sqrt{17}}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{6;-1;\dfrac{5-3\sqrt{17}}{2};\dfrac{5+3\sqrt{17}}{2}\right\}\)

`2)(6x+5)^2(3x+2)(x+1)=35`
`<=>12(6x+5)^2(3x+2)(x+1)=420`
`<=>(6x+5)^2+(6x+4)(6x+6)=420`
Đặt `6x+5=a` 
`pt<=>a^2(a+1)(a-1)=420`
`<=>a^2(a^2-1)-420=0`
`<=>a^4-a^2-420=0`
`<=>` $\left[ \begin{array}{l}a^2=-20(False)\\a^2=21(True)\end{array} \right.$
`<=>` $\left[ \begin{array}{l}a=\sqrt{20}\\a=-\sqrt{20}\end{array} \right.$
`<=>` $\left[ \begin{array}{l}6x+5=\sqrt{20}\\6x+5=-\sqrt{20}\end{array} \right.$
`<=>` $\left[ \begin{array}{l}x=\dfrac{\sqrt{20}-5}{6}\\x=\dfrac{-\sqrt{20}-5}{6}\end{array} \right.$
Vậy `S={\frac{\sqrt{20}-5}{6},\frac{-\sqrt{20}-5}{6}}`