bai toan nay kho qua

a) \(\Rightarrow\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)^2=\dfrac{4}{25}\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{2}{5}\\\dfrac{1}{2}x-\dfrac{1}{3}=-\dfrac{2}{5}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{22}{15}\\x=-\dfrac{2}{15}\end{matrix}\right.\)

b) \(\Rightarrow\left(1-\dfrac{1}{4}x\right)^2=\dfrac{121}{49}\)

\(\Rightarrow\left[{}\begin{matrix}1-\dfrac{1}{4}x=\dfrac{11}{7}\\1-\dfrac{1}{4}x=-\dfrac{11}{7}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{16}{7}\\x=\dfrac{72}{7}\end{matrix}\right.\)

\(a,121-\left(115+x\right)=3x-\left(25-9-5x\right)-8\\ 121-115-x=3x-25+9+5x-8\\ 6-x=8x-24\\ 8x+x=-24-6\\ 9x=-30\\ x=-\dfrac{30}{9}=-\dfrac{10}{3}\\ ----\\ b,2^{x+2}.3^{x+1}.5^x=10800\\ \left(2.3.5\right)^x.2^2.3=10800\\ 30^x.12=10800\\ 30^x=\dfrac{10800}{12}=900=30^2\\ Vậy:x=2\)

a: Thay x=-3 vào A, ta được:

\(A=\dfrac{-3-5}{-3-4}=\dfrac{8}{7}\)

b: \(B=\dfrac{2}{x+5}+\dfrac{x+25}{\left(x+5\right)\left(x-5\right)}=\dfrac{2x-10+x+25}{\left(x+5\right)\left(x-5\right)}=\dfrac{3x+15}{\left(x-5\right)\left(x+5\right)}=\dfrac{3}{x-5}\)

c: Để M là số nguyên thì \(x-4\in\left\{1;-1;3;-3\right\}\)

hay \(x\in\left\{3;7;1\right\}\)

Bài 3:

Tổng số tiền An dùng mua đồ:

125 000 + 85 000 + 60 000 + 65 000 = 335 000 (đồng)

Số tiền An còn lại sau khi mua đồ:

350 000 - 335 000 = 15 000 (đồng)

Đ.số: 15 000 đồng

Bài 2:

a, IV: Bốn(4); XXVII: Hai mươi bảy (27), XXX: ba mươi (30), M:một nghìn (1000)

b, 7: VII; 15: XV; 29: XXIX

a ) \(\left(4x+5\right)\div3-121\div11=4\)

\(\left(4x+5\right)\div3-11=4\)

\(\left(4x+5\right)\div3=4+11\)

\(\left(4x+5\right)\div3=15\)

\(\left(4x+5\right)=15\cdot3\)

\(4x+5=45\)

\(4x=45-5\)

\(4x=40\)

\(x=10\)

(4x + 5) : 3 - 121 : 11 = 4

=> (4x + 5) : 3 - 11 = 4

=> (4x + 5) : 3 = 15

=> 4x + 5 = 45

=> 4x = 40

=> x = 10

b) 1 + 3 + 5 + ... + x = 1600

=>[(x - 1) : 2 + 1] . (x + 1) : 2 = 1600

=> \(\left(\frac{x}{2}-\frac{1}{2}+1\right).\frac{x+1}{2}=1600\)

=> \(\frac{x+1}{2}.\frac{x+1}{2}=1600\)

=> \(\left(\frac{x+1}{2}\right)^2=1600\)

=> \(\frac{x+1}{2}=40\)

=> x + 1 = 80

=> x = 79

a,\(\left(4x+5\right):3-121:11=4\)

\(< =>\frac{4x+5}{3}-11=4\)

\(< =>\frac{4x+5}{3}=4+11=15\)

\(< =>4x+5=15.3=45\)

\(< =>4x=45-5=40< =>x=10\)

b, \(1+3+5+...+x=1600\)

Số số hạng : \(\left(x-1\right):2+1\)

Tổng : \(\frac{\left(x+1\right)\left(\frac{x-1}{2}+1\right)}{2}=\frac{\left(x+1\right)\left(\frac{x+1}{2}\right)}{2}=1600\)

\(< =>\frac{\left(x+1\right)^2}{2}=1600.2=3200\)

\(< =>\left(x+1\right)^2=3200.2=6400\)

\(< =>\left(x+1\right)^2=80^2=\left(-80\right)^2\)

\(< =>\orbr{\begin{cases}x+1=80\\x+1=-80\end{cases}< =>\orbr{\begin{cases}x=79\left(thoa-man\right)\\x=-81\left(loai\right)\end{cases}}}\)(do x > 0)

1. So sánh

a) \(25^{50}\) và \(2^{300}\)

\(25^{50}=25^{1.50}=\left(25^1\right)^{50}=25^{50}\)

\(2^{300}=2^{6.50}=\left(2^6\right)^{50}=64^{50}\)

Vì \(25< 64\) nên \(25^{50}< 64^{50}\)

Vậy \(25^{50}< 2^{300}\)

b) \(625^{15}\) và \(12^{45}\)

\(625^{15}=625^{1.15}=\left(625^1\right)^{15}=625^{15}\)

\(12^{45}=12^{3.15}=\left(12^3\right)^{15}=1728^{15}\)

Vì \(625< 1728\) nên \(625^{15}< 1728^{15}\)

Vậy \(625^{15}< 12^{45}\)

1.So sánh

a)\(25^{50}\) và \(2^{300}\)

Ta có : \(2^{300}=\left(2^6\right)^{50}=64^{50}\)

Vì \(25^{50}< 64^{50}\) nên \(25^{50}< 2^{300}\)

b)\(625^{15}\) và \(12^{45}\)

Ta có : \(12^{45}=\left(12^3\right)^{15}=1728^{15}\)

Vì \(625^{15}< 1728^{15}\) nên \(625^{15}< 12^{45}\)

(x-1)⁵=1

(x-1)⁵=1⁵

x-1=1

x=1+1

x=2

(3x-1)²=121

(3x-1)²=11²

3x-1=11

3x=11+1

3x=12

x=12:3

x=4

4^2x+1=64

4^2x+1=4³

=>2x+1=3

=>2x=3-1

=>2x=2

=>x=2:2

=>x=1

(x-2)⁵=32

(x-2)⁵=2⁵

x-2=2

x=2+2

x=4

(2^3-2^3)-(2^2-(-2)^2+(4^2*(-4)^2)=0-0+256=256

2, tìm x thuộc Z biết :

a, x^2 -(-3 )^2 =16

x^2-9              = 16

x^2                  = 25

=> x = 5

b, x^2 + (-4) ^2 =0 

x^2      + 16       = 0

x^2                    = -16

=> x=  -4