Tuyển Cộng tác viên Hoc24 nhiệm kì 26 tại đây: https://forms.gle/dK3zGK3LHFrgvTkJ6
ta có:
\(A=\dfrac{xy}{xy}-\dfrac{x-y}{y-x}.\left(\dfrac{x}{x}-\dfrac{y}{y}\right)\\ =1-\dfrac{x-y}{y-x}.\left(1-1\right)\\ =1-\dfrac{x-y}{y-x}.0\\ =1-0\\ =1\)
Tính giá trị biểu thức
\(A=\dfrac{\left(a+b\right)\left(-x-y\right)-\left(a-y\right)\left(b-c\right)}{abxy\left(xy+ay+ab+bx\right)}\) với \(a=-2;x=\dfrac{1}{3};b=\dfrac{2}{3};y=-1\)
tìm x;y
b \(\left|x-y\right|+\left|y+\dfrac{9}{25}=0\right|\)
c \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}=0\)
a \(\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|=\dfrac{-1}{4}-y\)
d \(\left|x\left(x^2-\dfrac{5}{4}\right)\right|=x\) \(\left(x\ge0\right)\)
e \(x^2+\left(y-\dfrac{1}{10}\right)^4=0\)
a, cho a, b là 2 số thoả mãn |a-2b+3|\(^{2023}\) + (b-1)\(^{2024}\) = 0. Tính giá trị biểu thức
P = a\(^{2023}\) x b\(^{2024}\) + 2024
b, 3 số hữu tỉ x,y,z thoả mãn xy+yz+zx = 2023. Chứng tỏ rằng:
A = \(\dfrac{\left(x^2+2023\right)x\left(y^2+2023\right)x\left(z^2+2023\right)}{16}\) viết được dưới dạng bình phương của 1 số hữu tỉ
Bài 4: Cho \(\dfrac{x+y-z}{x}=\dfrac{y+z-x}{y}=\dfrac{z+x-y}{z}\)
Tính A = \(\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)\)
Tìm x, y, z:
a) \(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2005\right|=0\)
b) \(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|\) = 0
c) \(\dfrac{16}{2^x}=1\)
d) \(\left(2x-1\right)^3=-27\)
e) \(\left(x-2\right)^2=1\)
f) \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{4}{25}\)
g) \(\left(x-1\right)^2=\left(x-1\right)^6\)
cho a(y+z)=b(z+x)=c(x+y) chứng minh \(\dfrac{y-z}{a\left(b-c\right)}=\dfrac{z-x}{b\left(c-a\right)}=\dfrac{x-y}{c\left(a-b\right)}\)
\(\dfrac{x-y+z}{z}=\dfrac{y+z-x}{x}=\dfrac{x-y+z}{y}\)
Tính A= \(\left(1+\dfrac{y}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{x}{z}\right)\)
Tìm x,y,z
\(\left|x-\dfrac{1}{2}\right|+\left|y+\dfrac{2}{3}\right|+\left|x^2+xz\right|=0\)
Cho : \(\dfrac{x^2-yz}{a}=\dfrac{y^2-zx}{b}=\dfrac{z^2-xy}{c}\)
CM : \(\dfrac{a^2-bc}{x}=\dfrac{b^2-ac}{y}=\dfrac{c^2-ab}{z}\)