a) 64/169=(-8/13)x
b)9x:3x=3
Rút gọn biểu thức : a . A = 4 √25x/4 - 8/3 √9x/4 - 4/3x √9x³/64 ( với x ≥ 0 ) b. B = y/2 + 3/4 √1-4y+4y² - 3/2 ( với y ≤ 1/2 )
a: \(A=4\cdot\dfrac{5}{2}\sqrt{x}-\dfrac{8}{3}\cdot\dfrac{3}{2}\sqrt{x}-\dfrac{4}{3x}\cdot\dfrac{3x}{8}\cdot\sqrt{x}\)
\(=10\sqrt{x}-4\sqrt{x}-\dfrac{1}{2}\sqrt{x}\)
\(=\dfrac{11}{2}\sqrt{x}\)
b: \(B=\dfrac{y}{2}+\dfrac{3}{4}\cdot\left|2y-1\right|-\dfrac{3}{2}\)
\(=\dfrac{y}{2}+\dfrac{3}{4}\left(1-2y\right)-\dfrac{3}{2}\)
=1/2y+3/4-3/2y-3/2
=-y-3/4
Rút gọn biểu thức :
a) A = 4\(\sqrt{\frac{25x}{4}}-\frac{8}{3}\sqrt{\frac{9x}{4}}-\frac{4}{3x}\sqrt{\frac{9x^3}{64}}\)với x > 0
Lời giải:
$A=4.\sqrt{\frac{25}{4}}.\sqrt{x}-\frac{8}{3}.\sqrt{\frac{9}{4}}.\sqrt{x}-\frac{4}{3x}.\sqrt{\frac{9}{64}}.\sqrt{x^3}$
$=10\frac{x}-4\sqrt{x}-\frac{1}{2x}.x\sqrt{x}=10x-4x-\frac{1}{2}\sqrt{x}$
$=\frac{11}{2}\sqrt{x}$
rút gọn biểu thức
A=\(4\sqrt{\dfrac{25x}{4}}-\dfrac{8}{3}\sqrt{\dfrac{9x}{4}}-\dfrac{4}{3x}\sqrt{\dfrac{9x^3}{64}}\) với \(x>0\)
\(A=4\sqrt{\dfrac{25x}{4}}-\dfrac{8}{3}\sqrt{\dfrac{9x}{4}}-\dfrac{4}{3x}\sqrt{\dfrac{9x^3}{64}}\)
\(A=4\left|\dfrac{5\sqrt{x}}{2}\right|-\dfrac{8}{3}\left|\dfrac{3\sqrt{x}}{2}\right|-\dfrac{4}{3x}\left|\dfrac{3x\sqrt{x}}{8}\right|\)
Vì \(x>0\) nên:
\(A=10\sqrt{x}-4\sqrt{x}-\dfrac{1}{2}\sqrt{x}=\dfrac{11\sqrt{x}}{2}\)
Tìm số nguyên x
\(\frac{64}{169}=\left(-\frac{8}{13}\right)^x\)
minh giai nhu the nay xem dung ko nha:
x se la mu 2 :
nhu vay ta se duoc :
\(\left(-8\right)^2\) = 64 { vi (- 8) . (- 8) = 64 vi am nham voi am se ra duong }
tuong tu : \(13^2\)= 169 ( vi 13 .13 = 169 )
nen ta co \(\frac{64}{169}=\frac{64}{169}\)
nen ta => so nguyen x con co the noi la mu x = 2
vay \(\frac{64}{169}=\left(-\frac{8}{13}\right)^2\)
Bài 1 : Tính
a) ( x + 1 ) \(^2\)+ ( x - 2 ) ( x + 3 ) - 4 x
b) ( 6x\(^5\)y\(^2\)-9x\(^4\)y\(3\)+ 12x\(^3\)y\(^4\)) : 3x\(^3\)y\(2\)
a: \(\left(x+1\right)^2+\left(x+3\right)\left(x-2\right)-4x\)
\(=x^2+2x+1+x^2+x-6-4x\)
\(=2x^2-x-6\)
Rút gọn biểu thức
A=\(4\sqrt{\frac{25x}{4}}-\frac{8}{3}\sqrt{\frac{9x}{4}}-\frac{4}{3x}\sqrt{\frac{9x^3}{64}}\) với \(x\ge0\)
tính tổng sau
2+4+6+8+..........+150
7+10+13+........+76
49+64+81+.........+169
\(2+4+6+8+...+150=\frac{\left(150+2\right)\left[\left(150-2\right)\div2+1\right]}{2}\)
\(=\frac{152.75}{2}=76.75=5700\)
\(7+10+13+...+76=\frac{\left(76+7\right)\left[\left(76-7\right)\div3+1\right]}{2}\)
\(=\frac{83.24}{2}=83.12=996\)
\(49+64+81+...+169=7^2+8^2+9^2+...+13^2=584\)
a)\(\sqrt{4-5x}=12\) tìm x
b)\(\sqrt{10+\sqrt{3x}}=2+\sqrt{6}\)
c)\(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)
a) Ta có: \(\sqrt{4-5x}=12\)
\(\Leftrightarrow4-5x=144\)
\(\Leftrightarrow5x=-140\)
hay x=-28
b) Ta có: \(\sqrt{10+\sqrt{3x}}=2+\sqrt{6}\)
\(\Leftrightarrow\sqrt{3x}+10=10+4\sqrt{6}\)
\(\Leftrightarrow\sqrt{3x}=4\sqrt{6}\)
\(\Leftrightarrow3x=96\)
hay x=32
c) Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+\dfrac{4}{3}\cdot3\sqrt{x+5}=6\)
\(\Leftrightarrow x+5=4\)
hay x=-1
Rút gọn phân thức:
a) xy-3y-9x+27/3-x
b) (3x+2)^2-(x+2)^2/x^3-x^2
b: \(\dfrac{\left(3x+2\right)^2-\left(x+2\right)^2}{x^3-x^2}\)
\(=\dfrac{\left(3x+2+x+2\right)\left(3x+2-x-2\right)}{x^2\left(x-1\right)}\)
\(=\dfrac{\left(4x+4\right)\cdot2x}{x^2\left(x-1\right)}\)
\(=\dfrac{8x\left(x+1\right)}{x^2\left(x-1\right)}=\dfrac{8}{x}\)