Tìm x biết :
A.1 - 1,2 + x = 0,5
B.1,35 - x + 0,49 = 0,48
Các ae tốt ơi
Giúp tau với
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\(x\times0,49+x\times0,01=1,2\\ =>x\times\left(0,49+0,01\right)=1,2\\ =>x\times0,5=1,2\\ =>x=1,2:0,5\\ =>x=2,4\)
X x 0,49 + X x 0,01 = 1,2
X x (0,49 + 0,01) = 1,2
X x 0,5 = 1,2
X = 1,2 : 0,5
X = 2,4
\(1,35-x+0,49=0,48\)
\(x-0,49=1,35-0,48\)
\(x-0,49=0,87\)
\(x=0,87+0,49\)
\(x=1,36\)
Vậy x = 1,36
x2 (x - y) = 5 (y - 1)
=> x2 (x - y) (y - 1) = 5
=> x2 .xy - x - y - y = 5
=> x2 (xy - x) - 2y = 5
=> x2 - x(y + 1) - 2y = 5
=> (x2 - x) (y + 1) - 2y = 5
=> x(x - 1)(y + 1) - 2y = 5
=> (x - 1) (y + 1) (x - 2y) = 5
=> \(\hept{\begin{cases}x-1=5\\y+1=5\\x-2y=5\end{cases}}\)
bó tay1!!!!!!!!!!!!5645775687697897894525256346436546457567567576876887956867876
X x 1,2 + X x 1,8 = 1,35
X x ( 1,2 + 1,8 ) = 1,35
X x 3 = 1,35
X = 1,35 : 3
X = 0,45
Vậy x = 0,45
\(\frac{4,95}{x}=\frac{1,35}{x}+1,2\)
\(\Rightarrow\frac{99}{20x}=\frac{27}{20x}+\frac{6}{5}\)
\(\Rightarrow\frac{99}{20x}-\frac{27}{20x}=\frac{6}{5}\)
\(\Rightarrow\frac{72}{20x}=\frac{6}{5}\)
\(\Rightarrow\frac{18}{5x}=\frac{6}{5}\)
\(\Rightarrow x=\frac{18}{6}=3\)
Vậy x = 3
dễ thấy vế trái luôn>0 nên 6x>0=> x>0
x>0, bỏ dấu trị tuyệt đối ra ta đc 4x+10=6x
x=5
chúc bạn học giỏi, ăn Tết đc ngon, hehe -_-
HYC-30/1/2022
Answer:
\(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+\left|x+4\right|=6x\)
Có \(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+\left|x+4\right|\ge0\)
\(\Rightarrow6x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow x+1+x+2+x+3+x+4=6x\)
\(\Rightarrow4x+10=6x\)
\(\Rightarrow2x=10\)
\(\Rightarrow x=5\)
1) \(5-\left(1+\dfrac{1}{3}\right):\left(1-\dfrac{1}{3}\right)\)
\(=5-\dfrac{4}{3}:\dfrac{2}{3}\)
\(=5-\dfrac{4}{3}\cdot\dfrac{3}{2}\)
\(=5-\dfrac{4}{2}\)
\(=5-2\)
\(=3\)
b) \(\left(1+\dfrac{2}{3}-\dfrac{5}{4}\right)-\left(1-\dfrac{5}{4}\right)+2022-\dfrac{2}{3}\)
\(=1+\dfrac{2}{3}-\dfrac{5}{4}-1+\dfrac{5}{4}++2022-\dfrac{2}{3}\)
\(=\left(1-1\right)+\left(\dfrac{2}{3}-\dfrac{2}{3}\right)+\left(-\dfrac{5}{4}+\dfrac{5}{4}\right)+2022\)
\(=0+0+0+2022\)
\(=2022\)
2) \(0,7^2\cdot x=0,49^2\)
\(\Rightarrow x=\dfrac{0,49^2}{0,7^2}\)
\(\Rightarrow x=\left(\dfrac{0,49}{0,7}\right)^2\)
\(\Rightarrow x=\left(0,7\right)^2\)
\(\Rightarrow x=0,49\)
b) \(x:\left(-0,5\right)^3=\left(0,5\right)^2\)
\(\Rightarrow x=\left(0,5\right)^2\cdot\left(-0,5\right)^3\)
\(\Rightarrow x=\left(-0,5\right)^5\)
\(\Rightarrow x=-\dfrac{1}{32}\)
2:
a: =>x*0,49=0,49^2
=>x=0,49
b: =>x=(0,5)^2*(-1)*(0,5)^3=-(0,5)^5
b. \(\dfrac{13+x}{20}\)= \(\dfrac{15}{20}\)
=> 13+x=15
x=15-13
x=2
$#Shả$
`x xx1,2+x xx1,8=45`
`<=>x xx(1,2+1,8)=45`
`<=> x xx 3 =45`
`<=>x=45:3=15`
`(13+x)/20=3/4`
`<=>4xx(13+x)=20xx3`
`<=>4xx(13+x)=60`
`<=>13+x=60:4=15`
`<=>x=15-13=2`
a, 1,2+x=1-0,5
1,2+x=0,5
x=0,5-1,2
x=-0,7
b, x+0,49=1,35-0,48
x+0,49=0,87
x= 0,87-0,49
x=0,38
k mik nha
a, \(1-1,2+x=0,5\)
\(\Rightarrow-0,2+x=0,5\)
\(\Rightarrow x=0,5-(-0,2)=0,5+0,2=0,7\)
b,\(1,35-x+0,49=0,48\)
\(\Rightarrow1,35-x=0,48-0,49\)
\(\Rightarrow1,35-x=-0,01\)
\(\Rightarrow x=1,35-(-0,01)=1,35+0,01=1,36\)