BT1: Tìm x, biết:
2) \(9-\left(\dfrac{2}{3}-x\right)=-\dfrac{5}{4}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
DH
24 tháng 7 2017
\(-\dfrac{2}{5}+\dfrac{5}{3}\left(\dfrac{3}{2}-\dfrac{4}{15}x\right)=-\dfrac{7}{6}\)
\(\Rightarrow\dfrac{5}{3}\left(\dfrac{3}{2}-\dfrac{4}{15}x\right)=-\dfrac{23}{30}\)
\(\Rightarrow\dfrac{3}{2}-\dfrac{4}{15}x=-\dfrac{23}{50}\)
\(\Rightarrow\dfrac{4}{15}x=\dfrac{49}{25}\Rightarrow x=\dfrac{147}{20}\)
Chúc bạn học tốt!!!
28 tháng 9 2021
a) \(\left|3x-\dfrac{1}{2}\right|+\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|=0\)
Do \(\left|3x-\dfrac{1}{2}\right|,\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|\ge0\forall x,y\)
\(\Rightarrow\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\\dfrac{1}{4}y+\dfrac{3}{5}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-\dfrac{12}{5}\end{matrix}\right.\)
b) \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|+\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\le0\)
Do \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|,\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\ge0\forall x,y\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{9}=0\\\dfrac{5}{7}y-\dfrac{1}{2}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{27}\\y=\dfrac{7}{10}\end{matrix}\right.\)
H9
HT.Phong (9A5)
CTVHS
1 tháng 11 2023
a) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\left(-\dfrac{7}{12}\right)\cdot1\dfrac{2}{5}\)
\(\Rightarrow\dfrac{1}{6}x=\left(-\dfrac{7}{12}\right)\cdot\dfrac{7}{5}\)
\(\Rightarrow\dfrac{1}{6}x=-\dfrac{49}{60}\)
\(\Rightarrow x=-\dfrac{49}{60}:\dfrac{1}{6}\)
\(\Rightarrow x=-\dfrac{49}{10}\)
b) \(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\dfrac{9}{4}\)
\(\Rightarrow\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\left(\pm\dfrac{3}{2}\right)^2\)
+) \(\dfrac{1}{5}-\dfrac{3}{2}x=\dfrac{3}{2}\)
\(\Rightarrow\dfrac{3}{2}x=\dfrac{1}{5}-\dfrac{3}{2}\)
\(\Rightarrow\dfrac{3}{2}x=-\dfrac{13}{10}\)
\(\Rightarrow x=-\dfrac{13}{10}:\dfrac{3}{2}\)
\(\Rightarrow x=-\dfrac{13}{15}\)
+) \(\left(1,25-\dfrac{4}{5}x\right)^3=-125\)
\(\Rightarrow\left(\dfrac{5}{4}-\dfrac{4}{5}x\right)^3=\left(-5\right)^3\)
\(\Rightarrow\dfrac{5}{4}-\dfrac{4}{5}x=-5\)
\(\Rightarrow\dfrac{4}{5}x=\dfrac{5}{4}+5\)
\(\Rightarrow\dfrac{4}{5}x=\dfrac{25}{4}\)
\(\Rightarrow x=\dfrac{25}{4}:\dfrac{4}{5}\)
\(\Rightarrow x=\dfrac{125}{16}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
1 tháng 11 2023
a, \(\dfrac{2}{3}\)\(x\) - \(\dfrac{1}{2}\)\(x\) = (- \(\dfrac{7}{12}\)). 1\(\dfrac{2}{5}\)
\(x\).(\(\dfrac{2}{3}\) - \(\dfrac{1}{2}\)) = (- \(\dfrac{7}{12}\)) . \(\dfrac{7}{5}\)
\(x\). \(\dfrac{1}{6}\) = - \(\dfrac{49}{60}\)
\(x\) = - \(\dfrac{49}{60}\).6
\(x\) = -\(\dfrac{49}{10}\)
H9
HT.Phong (9A5)
CTVHS
11 tháng 7 2023
a) \(\left(\dfrac{x^2}{2}+y^2\right)^2\)
\(=\left(\dfrac{1}{2}x^2+y^2\right)^2\)
\(=\left(\dfrac{1}{2}x^2\right)^2+2\cdot\dfrac{1}{2}x^2\cdot y^2+\left(y^2\right)^2\)
\(=\dfrac{1}{4}x^4+x^2y^2+y^4\)
b) \(\left(\dfrac{4}{5}x^2-\dfrac{2}{3}y\right)^2\)
\(=\left(\dfrac{4}{5}x^2\right)^2-2\cdot\dfrac{4}{5}x^2\cdot\dfrac{2}{3}y+\left(\dfrac{2}{3}y\right)^2\)
\(=\dfrac{16}{25}x^4-\dfrac{16}{15}x^2y+\dfrac{4}{9}y^2\)
c) \(\left(2x+\dfrac{1}{2}\right)\left(2x-\dfrac{1}{2}\right)\)
\(=\left(2x\right)^2-\left(\dfrac{1}{2}\right)^2\)
\(=4x^2-\dfrac{1}{4}\)
10 tháng 7 2023
a: (1/2x^2+y^2)^2
=(1/2x^2)^2+2*1/2x^2*y^2+y^4
=1/4x^4+x^2y^2+y^4
b: (4/5x^2-2/3y)^2
=(4/5x^2)^2-2*4/5x^2*2/3y+4/9y^2
=16/25x^4-16/15x^2y+4/9y^2
c: =(2x)^2-(1/2)^2
=4x^2-1/4
H9
HT.Phong (9A5)
CTVHS
11 tháng 7 2023
d) \(\left(x+2\right)\left(x^2-2x+4\right)\)
\(=\left(x+2\right)\left(x^2-2\cdot x+2^2\right)\)
\(=x^3+2^3\)
\(=x^3+8\)
e) \(\left(\dfrac{1}{4}-\dfrac{x}{5}\right)\left(\dfrac{x^2}{25}+\dfrac{x}{20}+\dfrac{1}{16}\right)\)
\(=\left(\dfrac{1}{4}-\dfrac{1}{5}x\right)\left(\dfrac{1}{25}x^2+\dfrac{1}{5}x\cdot\dfrac{1}{4}+\dfrac{1}{16}\right)\)
\(=\left(\dfrac{1}{4}-\dfrac{1}{5}x\right)\left[\left(\dfrac{1}{5}x\right)^2+\dfrac{1}{5}x\cdot\dfrac{1}{4}+\left(\dfrac{1}{4}\right)^2\right]\)
\(=\left(\dfrac{1}{4}\right)^3-\left(\dfrac{1}{5}x\right)^3\)
\(=\dfrac{1}{64}-\dfrac{1}{125}x^3\)
\(=\dfrac{1}{64}-\dfrac{x^3}{125}\)
10 tháng 7 2023
d: (x+2)(x^2-2x+4)
=(x+2)(x^2-x*2+2^2)
=x^3+8
e: (1/4-x/5)(1/16+x/20+x^2/25)
=(1/4-x/5)[(1/4)^2+1/4*x/5+(x/5)^2]
=1/64-x^3/125
KL
1
KL
1 tháng 8 2017
(x + \(\dfrac{3}{5}\))2 + 1\(\dfrac{16}{25}=9\%:4,5\%\)
<=> x2 + (\(\dfrac{3}{5}\))2 + 1\(\dfrac{16}{25}=2\)
<=> x2 + \(\dfrac{9}{25}+\dfrac{41}{25}=2\)
<=> x2 + 2 = 2
<=> x2 = 0
<=> x = 0
@Khánh Linh
Ta có:
9 - (2/3 - x) = -5/4
2/3 - x = 9 - (-5/4)
2/3 - x = 9 + 5/4
2/3 + x = 41/4
x = 41/4 - 2/3
x = 123/12 - 8/12
x = 115/12