Câu 1 The function mm is defined on the real numbers by m(k) = \dfrac{k+2}{k+8}m(k)= k+8 k+2 ​ . What is the value of 10\times m(2)10×m(2)? Answer: Câu 2 The function ff is defined on the real numbers by f(x)= ax-3f(x)=ax−3. What is the value of a if f(3)=9f(3)=9? Answer: Câu 3 The function ff is defined on the real numbers by f(x)= 2x+a-3f(x)=2x+a−3. What is the value of a if f(-5)=11f(−5)=11? Answer: Câu 4 The function ff is defined on the real numbers by f(x) = 2 + x-x^2f(x)=2+x−x 2 . What is the value of f(-3)f(−3)? Answer: Câu 5 Given a real number aa and a function ff is defined on the real numbers by f(x)=-6\times|3x|-4f(x)=−6×∣3x∣−4. Compare: f(a)f(a) f(-a)f(−a) Câu 6 There are ordered pairs (x;y)(x;y) where xx and yy are integers such that \dfrac{5}{x}+\dfrac{y}{4}=\dfrac{1}{8} x 5 ​ + 4 y ​ = 8 1 ​ Câu 7 Given a negative number kk and a function ff is defined on the real numbers by f(x)=\dfrac{6}{13}xf(x)= 13 6 ​ x. Compare: f(k)f(k) f(-k)f(−k) Câu 8 Given a positive number kk and a function ff is defined on the real numbers by f(x)=\dfrac{-3}{4}x+4f(x)= 4 −3 ​ x+4. Compare: f(k)f(k) f(-k)f(−k). Câu 9 A=(1+2+3+\ldots+90) \times(12 \times34-6 \times 68):(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6})=A=(1+2+3+…+90)×(12×34−6×68):( 3 1 ​ + 4 1 ​ + 5 1 ​ + 6 1 ​ )= Câu 10 Given that \dfrac{2x+y+z+t}{x}=\dfrac{x+2y+z+t}{y}=\dfrac{x+y+2z+t}{z}=\dfrac{x+y+z+2t}{t} x 2x+y+z+t ​ = y x+2y+z+t ​ = z x+y+2z+t ​ = t x+y+z+2t ​ . The negative value of \dfrac{x+y}{z+t}+\dfrac{y+z}{t+x}+\dfrac{z+t}{x+y}+\dfrac{t+x}{y+z} z+t x+y ​ + t+x y+z ​ + x+y z+t ​ + y+z t+x ​ is

Tìm x, y, z

\(\dfrac{x+y+2}{z}=\dfrac{y+z+1}{x}=\dfrac{z+x-3}{y}=\dfrac{1}{x+y+z}\)

Áp dụng tích chất của dãy tỉ số bằng nhau, ta có

\(\dfrac{x+y+2}{z}=\dfrac{y+z+1}{x}=\dfrac{z+x-3}{y}\\ =\dfrac{x+y+2+y+z+1+z+x-3}{z+x+y}=\dfrac{2\left(x+y+z\right)+\left(1+2-3\right)}{z+x+y}=2\\ Vì\dfrac{x+y+2}{z}=\dfrac{y+z+1}{x}=\dfrac{z+x-3}{y}=\dfrac{1}{x+y+z}\\ =>2=\dfrac{1}{x+y+z}=>2\left(x+y+z\right)=1=>x+y+z=\dfrac{1}{2}\\ =>\dfrac{x+y+2}{z}=2=>x+y+2=2z\\ \dfrac{y+z+1}{x}=2=>y+z+1=2x\\ \dfrac{z+x-3}{y}=2=>z+x-3=2y\\ \dfrac{1}{x+y+z}=2=>x+y+z=\dfrac{1}{2}\)

+) x+y+z = \(\dfrac{1}{2}=>y+z=\dfrac{1}{2}-x=>\dfrac{1}{2}-x+1=2x=>3x=\dfrac{3}{2}=>x=\dfrac{1}{2}\)

+)\(x+y+z=\dfrac{1}{2}=>x+y=\dfrac{1}{2}-z=>\dfrac{1}{2}-z+2=2z=>3z=\dfrac{5}{2}=>z=\dfrac{5}{6}\)

\(=>x+y+z=\dfrac{1}{2}+\dfrac{5}{6}+y=\dfrac{1}{2}=>\dfrac{4}{3}+y=\dfrac{1}{2}=>y=\dfrac{-5}{6}\)

Vậy \(x=\dfrac{1}{2}\\ y=\dfrac{-5}{6}\\ z=\dfrac{5}{6}\)

Ê mấy bọn 7B Nguyễn Lương Bằng ơi bài 2 Toán chiều làm thế này đúng chưa! Góp ý nha!