Ta có \(5^5-5^4+5^3=5^3\left(5^2-5+1\right)=5^3.21=5^3.3.7\)

Vì 5³.3 là số nguyên nên \(5^3.3.7⋮7\)

Vậy  \(5^5-5^4+5^3⋮7\)

c) \(3^{x+3}+3^{x+1}+2^{x+3}+2^{x+2}\)

\(=\left(3^{x+3}+3^{x+1}\right)+\left(2^{x+3}+2^{x+2}\right)\)

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

\(=3^x.12+2^x.6\)

\(=6\left(2.3^x+2^x\right)\)

Vì \(2.3^x+2^x\in Z\)

Nên : \(6\left(2.3^x+2^x\right)⋮6\)

Vậy \(3^{x+3}+3^{x+1}+2^{x+3}+2^{x+2}⋮6\)

a) ta có : \(5^5-5^4+5^3=5^3.\left(5^2-5+1\right)=5^3.\left(25-5+1\right)\)

\(5^3.21=5^3.3.7⋮7\) (đpcm)

b) ta có : \(7^6+7^5-7^4=7^4.\left(7^2+7-1\right)=7^4.\left(49+7-1\right)\)

\(=7^4.55=7^4.5.11⋮11\) (đpcm)

c) ta có : \(3^{x+2}-2^{x+3}+3^x-2^{x+1}=3^{x+2}+3^x-2^{x+3}-2^{x+1}\)

\(=3^x\left(3^2+1\right)-2^x\left(2^3+2\right)=3^x.\left(9+1\right)-2^x.\left(8+2\right)\)

\(=3^x.10-2^x.10=10\left(3^x-2^x\right)⋮10\) (đpcm)

d) \(3^{x+3}+3^{x+1}+2^{x+3}+2^{x+2}=3^x.\left(3^3+3\right)+2^x.\left(2^3+2^2\right)\)

\(=3^x.\left(27+3\right)+2^x\left(8+4\right)=3^x.30+2^x.12=6.\left(3^x.5+2^x.2\right)⋮6\) (đpcm)

a)Ta có:\(5^5-5^4+5^3=5^3\left(5^2-5+1\right)=5^3.21\)(vì 21 chia hết cho 7)

\(\)\(\RightarrowĐPCM\)

b)Ta có: \(7^6+7^5-7^4⋮11=7^4\left(7^2+7-1\right)=7^4.55⋮11\)

\(\Rightarrowđpcm\)

1)\(2x^2+9y^2-6xy-6x-12y+2004\)

\(=x^2+x^2-6xy+9y^2-6x-12y+2004\)

\(=x^2+\left(x-3y\right)^2-10x+4x-12y+2004\)

\(=\left(x-3y\right)^2+4\left(x-3y\right)+x^2-10x+2004\)

\(=\left(x-3y\right)^2+4\left(x-3y\right)+x^2-10x+4+25+1975\)

\(=\left[\left(x-3y\right)^2+4\left(x-3y\right)+4\right]+\left(x^2-10x+25\right)+1975\)

\(=\left(x-3y+2\right)^2+\left(x-5\right)^2+1975\ge1975\)

Dấu "=" khi \(\begin{cases}\left(x-5\right)^2=0\\\left(x-3y+2\right)^2=0\end{cases}\)\(\Leftrightarrow\begin{cases}x=5\\y=\frac{7}{3}\end{cases}\)

Vậy Min=1975 khi \(\begin{cases}x=5\\y=\frac{7}{3}\end{cases}\)

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

Đặt \(t=x^2+x\) ta có:

\(t\left(t-4\right)=t^2-4t+4-4\)

\(=\left(t-2\right)^2-4\ge-4\)

Dấu "=" khi \(t-2=0\Leftrightarrow t=2\Leftrightarrow x^2+x=2\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-2\\x=1\end{array}\right.\)

Vậy Min=-4 khi \(\left[\begin{array}{nghiempt}x=-2\\x=1\end{array}\right.\)

3)\(\left(x^2+5x+5\right)\left[\left(x+2\right)\left(x+3\right)+1\right]\)

\(=\left(x^2+5x+5\right)\left[x^2+5x+6+1\right]\)

Đặt \(t=x^2+5x+5\) ta có:

\(t\left(t+1\right)=t^2+t+\frac{1}{4}-\frac{1}{4}=\left(t+\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)

Dấu "=" khi \(t+\frac{1}{2}=0\Leftrightarrow t=-\frac{1}{2}\Leftrightarrow x^2+5x+5=-\frac{1}{2}\)\(\Leftrightarrow x_{1,2}=\frac{-10\pm\sqrt{12}}{4}\)

Vậy Min=\(-\frac{1}{4}\) khi \(x_{1,2}=\frac{-10\pm\sqrt{12}}{4}\)

4)\(\left(x-1\right)\left(x-3\right)\left(x^2-4x+5\right)\)

\(=\left(x^2-4x+3\right)\left(x^2-4x+5\right)\)

Đặt \(t=x^2-4x+3\) ta có:

\(t\left(t+2\right)=t^2+2t+1-1=\left(t+1\right)^2-1\ge-1\)

Dấu "=" khi \(t+1=0\Leftrightarrow t=-1\Leftrightarrow x^2-4x+3=-1\Leftrightarrow x=2\)

Vậy Min=-1 khi x=2

Thank you !

2, \(\Rightarrow\left\{{}\begin{matrix}\\\dfrac{5}{4}x-\dfrac{7}{2}=0\\\dfrac{5}{8}x+\dfrac{3}{5}=0\\\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{14}{5}\\\\x=\dfrac{-24}{25}\\\end{matrix}\right.\)

a, \(5^x+5^{x+1}+5^{x-2}=151\)

\(\Rightarrow5^x.\left(1+5+5^{-2}\right)=151\)

\(\Rightarrow5^x.6,04=151\Rightarrow5^x=25=5^2\)

Vì \(5\ne-1;5\ne0;5\ne1\) nên \(x=2\)

b, \(5^{x-1}+5^{x-2}+5^{x-3}=155\)

\(\Rightarrow5^x.\left(5^{-1}+5^{-2}+5^{-3}\right)=155\)

\(\Rightarrow5^x.0,248=155\Rightarrow5^x=625=5^4\)

Vì \(5\ne-1;5\ne0;5\ne1\) nên \(x=4\)

\(•5^x+5^{x+1}+5^{x-2}=151\\ 5^x\left(1+5+\dfrac{1}{25}\right)=151\\ 5^x=25\\ \Rightarrow x=2\)

\(•5^{x-1}+5^{x-2}+5^{x-3}=155\\ 5^x.\left(\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}\right)=155\\ 5^x=625\\ \Rightarrow x=4\)

\(•5^{2+x}+5^{3+x}=750\\ 5^x\left(25+125\right)=750\\ 5^x=5\\ \Rightarrow x=1\)

a) \(2\left(4x-30\right)-3\left(x+5\right)+4\left(x-10\right)=5\left(x+2\right)\)

\(\Leftrightarrow8x-60-3x+15+4x-40=5x+10\)

\(\Leftrightarrow9x-35=5x+10\)

\(\Leftrightarrow9x-5x=10+35\)

\(\Leftrightarrow4x=45\)

\(\Leftrightarrow x=\dfrac{45}{4}=11,25\)

b) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\left(6x+1\right)\)

\(\Leftrightarrow\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=4x+\dfrac{2}{3}\)

\(\Leftrightarrow\dfrac{31}{60}+x=4x+\dfrac{2}{3}\)

\(\Leftrightarrow\dfrac{31}{60}-\dfrac{2}{3}=4x-x\)

\(\Leftrightarrow3x=\dfrac{1}{60}\)

\(\Leftrightarrow x=\dfrac{1}{180}\)

c) \(\dfrac{7}{3}-\left(2x-\dfrac{1}{3}\right)=\left(-2\dfrac{1}{6}+1\dfrac{1}{2}\right):0,25\)

\(\Leftrightarrow\dfrac{7}{3}-2x+\dfrac{1}{3}=-1\dfrac{2}{3}:\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{8}{3}-2x=\dfrac{-5}{3}.4\)

\(\Leftrightarrow\dfrac{8}{3}-2x=\dfrac{-20}{3}\)

\(\Leftrightarrow2x=\dfrac{8}{3}+\dfrac{20}{3}\)

\(\Leftrightarrow2x=\dfrac{28}{3}\)

\(\Leftrightarrow x=4\dfrac{2}{3}\)

d) \(0,75+\dfrac{5}{9}:x=5\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{3}{4}+\dfrac{5}{9}:x=\dfrac{11}{2}\)

\(\Leftrightarrow\dfrac{5}{9}:x=\dfrac{11}{2}-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{5}{9}:x=\dfrac{19}{4}\)

\(\Leftrightarrow x=\dfrac{5}{9}:\dfrac{19}{4}\)

\(\Leftrightarrow x=\dfrac{20}{171}\)

a: \(\Leftrightarrow\dfrac{7}{2}x-\dfrac{3}{4}=\dfrac{1}{2}x+\dfrac{5}{2}\)

\(\Leftrightarrow3x=\dfrac{5}{2}+\dfrac{3}{4}=\dfrac{10}{4}+\dfrac{3}{4}=\dfrac{13}{4}\)

=>x=13/12

b: \(\Leftrightarrow x\cdot\left(\dfrac{2}{3}-\dfrac{1}{2}\right)=-\dfrac{1}{3}+\dfrac{2}{5}\)

\(\Leftrightarrow x\cdot\dfrac{1}{6}=\dfrac{-5+6}{15}=\dfrac{1}{15}\)

\(\Leftrightarrow x=\dfrac{1}{15}:\dfrac{1}{6}=\dfrac{2}{5}\)

c: \(\Leftrightarrow x\cdot\dfrac{1}{3}+x\cdot\dfrac{2}{5}+\dfrac{2}{5}=0\)

\(\Leftrightarrow x\cdot\dfrac{11}{15}=-\dfrac{2}{5}\)

\(\Leftrightarrow x=-\dfrac{2}{5}:\dfrac{11}{15}=\dfrac{-2}{5}\cdot\dfrac{15}{11}=\dfrac{-30}{55}=\dfrac{-6}{11}\)

d: \(\Leftrightarrow-\dfrac{1}{3}x+\dfrac{1}{2}+\dfrac{2}{3}-x-\dfrac{1}{2}=5\)

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

\(\Leftrightarrow-\dfrac{4}{3}x=5-\dfrac{2}{3}=\dfrac{13}{3}\)

\(\Leftrightarrow x=\dfrac{13}{3}:\dfrac{-4}{3}=\dfrac{-13}{4}\)

e: \(\Leftrightarrow\left(\dfrac{x+2015}{5}+1\right)+\left(\dfrac{x+2016}{4}+1\right)=\left(\dfrac{x+2017}{3}+1\right)+\left(\dfrac{x+2018}{2}+1\right)\)

=>x+2020=0

hay x=-2020

58. \(\left|x-\frac15\right|+\frac13=\frac14-\left|-\frac32\right|\)

\(\left|x-\frac15\right|+\frac13=\frac14-\frac32\)

\(\left|x-\frac15\right|=\frac14-\frac32-\frac13\)

\(\left|x-\frac15\right|=-\frac{19}{12}\)

⇒ vô nghiệm

59. \(\left|x-\frac52\right|=\frac43-\left(\frac23-\frac12\right)\)

\(\left|x-\frac52\right|=\frac76\)

\(\left[\begin{array}{l}x-\frac52=\frac76\Rightarrow x=\frac{11}{3}\\ x-\frac52=-\frac76\Rightarrow x=\frac43\end{array}\right.\)

vậy \(x\in\left\lbrace\frac43;\frac{11}{3}\right\rbrace\)

Để tìm x, ta giải từng phương trình:

Đối với phương trình 58:

|x - 1/5| + 1/3 = 1/4 - |-3/2|, ta biến đổi thành |x - 1/5| = 1/4 - 3/2 - 1/3 = -11/12.

Vì giá trị tuyệt đối luôn lớn hơn hoặc bằng 0, phương trình này không có nghiệm x.

Đối với phương trình 59:

|x - 5/2| = 4/3 - (2/3 - 1/2), ta biến đổi thành |x - 5/2| = 4/3 - (4/6 - 3/6) = 4/3 - 1/6 = 8/6 - 1/6 = 7/6. Suy ra x - 5/2 = 7/6 hoặc x - 5/2 = -7/6, cho ta các nghiệm x = 23/6 và x = 8/6 = 4/3.

Giải chi tiết:

Phương trình 58: |x - 1/5| + 1/3 = 1/4 - |-3/2| Tính giá trị tuyệt đối: |-3/2| = 3/2. Thay vào phương trình: |x - 1/5| + 1/3 = 1/4 - 3/2.

Chuyển 1/3 sang vế phải: |x - 1/5| = 1/4 - 3/2 - 1/3.

Quy đồng mẫu số để trừ: 1/4 - 6/4 - 4/12 = 3/12 - 18/12 - 4/12 = -19/12. Tuy nhiên, kiểm tra lại phép tính: 1/4 - 3/2 - 1/3 = 3/12 - 18/12 - 4/12 = (3 - 18 - 4)/12 = -19/12.

Nếu kết quả là -19/12 thì |x - 1/5| = -19/12. Vì giá trị tuyệt đối của một số luôn không âm, phương trình này không có nghiệm x.

Phương trình 59: |x - 5/2| = 4/3 - (2/3 - 1/2)

Tính biểu thức trong ngoặc: 2/3 - 1/2 = 4/6 - 3/6 = 1/6.

Thay vào phương trình: |x - 5/2| = 4/3 - 1/6.

Quy đồng mẫu số để trừ: |x - 5/2| = 8/6 - 1/6 = 7/6.

Trường hợp 1: x - 5/2 = 7/6.

x = 7/6 + 5/2.

x = 7/6 + 15/6.

x = 22/6 = 11/3.

Trường hợp 2: x - 5/2 = -7/6.

x = -7/6 + 5/2.

x = -7/6 + 15/6.

x = 8/6 = 4/3.

Kết luận:

Phương trình 58 không có nghiệm.

Phương trình 59 có hai nghiệm là x = 11/3 và x = 4/3.