Tìm x
\(\frac{x+1}{2x+1}=\frac{0.5x+2}{x+3}\)
Tìm x biết:
\(\frac{x+1}{2x+1}=\frac{0.5x+2}{x+3}\)
\(\Leftrightarrow\left(x+1\right)\left(x+3\right)=\left(2x+1\right)\left(x+3\right)\)
\(\Leftrightarrow x^2+4x+3=x^2+4.5x+2\)
\(\Leftrightarrow0.5x+2=3\Leftrightarrow0.5x=1\Leftrightarrow x=2\)
\(\frac{x+1}{2x+1}\)\(=\)\(\frac{0,5x+2}{x+3}\)
=> (x+1).(x+3) = (2x+1).(0,5x+2)
=> x2+4x+3 = x2+4,5x+2
=> x2+4x-x2-4,5x = 2-3
=> -0,5x = -1
=> x = -1:-0,5
=> x = 2
\(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}\)
\(\Leftrightarrow\left(x+1\right)\left(x+3\right)=\left(0,5x+2\right)\left(2x+1\right)\)
\(\Leftrightarrow x^2+3x+x+3=x^2+0,5x+4x+2\)
\(\Leftrightarrow x^2+4x+3=x^2+4,5x+2\)
\(\Leftrightarrow4x+3=4,5x+2\)
\(\Leftrightarrow0,5x=-1\)
\(\Leftrightarrow x=-2\)
Tíck cho mìk vs nha Vân Nguyễn!
tim x
\(\frac{3x+2}{5x+7}=\frac{3x-1}{5x+1}\)
\(\frac{x+1}{2x+1}=\frac{0.5x+2}{x+3}\)
\frac{2+x}{5}-0.5x=\frac{1-2x}{4}+0,\:25
a) \(\frac{3x+2}{5x+7}=\frac{3x-1}{5x+1}\)
b)\(\frac{x+1}{2x+1}=\frac{0.5x+2}{x+3}\)
a ) \(\frac{3x+2}{5x+7}=\frac{3x-1}{5x+1}\)
\(\Leftrightarrow\left(3x+2\right)\left(5x+1\right)=\left(3x-1\right)\left(5x+7\right)\)
\(\Leftrightarrow3x\left(5x+1\right)+2\left(5x+1\right)=3x\left(5x+7\right)-\left(5x+7\right)\)
\(\Leftrightarrow15x^2+3x+10x+2=15x^2+21x-5x-7\)
\(\Leftrightarrow15x^2+13x+2=15x^2+16x-7\)
\(\Leftrightarrow13x+2=16x-7\)
\(\Leftrightarrow13x-16x=-7-2\)
\(\Leftrightarrow-3x=-9\)
\(\Rightarrow x=3\)
b ) tương tự
tìm x biết:
a, \(\frac{1}{3}x-0.5x=0.75\)
\(\frac{1}{3}x-0,5x=0,75\)
\(\Leftrightarrow-\frac{1}{6}x=0,75\)
\(\Leftrightarrow\)\(x=-\frac{9}{2}\)
1/3x -0.5x = 0.75
(1/3 - 0.5)x = 0.75
-1/6x =0.75
x =0.75 : -1/6
x =-9/2
tk mình nha .thank bạn nhiều:))
x. (1/3 - 0,5) = 0,75
x. (-1/6) = 3/4
x = 3/4 : (-1/6)
x = -9/2
Tìm x biết:
1, 3x+2/5x+7=3x-1/5x+1
2, x+1/2x+1=0.5x+2/x+3
1,3x+2/5x+7 =3x-1/5x+1
<=> 1,3x+2/5x-3x+1/5x = 1-7
<=> (1,3+2/5-3+1/5)x = -6
<=> -11/10x=-6
<=> x= -6 : (-11/10)
<=> x= 60/11
2.x+1/2x+1 = 0,5x+2/x +3
<=> 2x+1/2x-0,5x-2/1x = 3-1
<=> x(2+1/2-0,5-2 ) =2
<=>0x =2
<=> x=0
Hinh nhu minh thay ban Kunzy Nguyen giai hoi sai
1,3x o dau ra ???????????????
Hình như bạn Kunzy Nguyễn ghi nhầm 3x thành 1,3x nhỉ ?!!!
Tìm x biết:
1, 3x+2 / 5x+7 = 3x-1 / 5x+1
2, x+1 / 2x+1 =0.5x+2 / x+3
1) áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{3x+2}{5x+7}=\frac{3x-1}{5x+1}=\frac{\left(3x+2\right)-\left(3x-1\right)}{\left(5x+7\right)-\left(5x+1\right)}=\frac{3x+2-3x+1}{5x+7-5x-1}=\frac{3}{6}=\frac{1}{2}\)
suy ra :
\(\frac{3x-1}{5x+1}=\frac{1}{2}\Rightarrow\left(5x+1\right).1=\left(3x-1\right).2\)
=> 5x+1=6x-2
5x-6x=-2-1
-x=-3
x=3
2)áp dụng tính chất của dãy tỉ số bằng nhau ta có;
\(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}=\frac{\left(x+1\right)-2.\left(0,5x+2\right)}{\left(2x+1\right)-2.\left(x+3\right)}=\frac{x+1-x-4}{2x+1-2x-6}=\frac{-3}{-5}=\frac{3}{5}\)
suy ra:
\(\frac{x+1}{2x+1}=\frac{3}{5}\Rightarrow\left(2x+1\right).3=\left(x+1\right).5\)
=>6x+3=5x+5
6x-5x=5-3
x=2
chỉ có làm mới có ăn
1) \(\frac{X+2}{X+3}+\frac{X-1}{X+1}=\frac{2}{X^2+4X+3}+1\)
2)\(\frac{X+1}{X-2}+\frac{2X-1}{X-1}=\frac{2}{X^2-3X+2}+\frac{11}{2}\)
3) Tìm GTLN CỦA -2X2+4X+3
4)\(\frac{X+1}{X-2}+\frac{X}{X+1}-\frac{2X+5}{X^2-X-2}=2\)
5)\(\frac{2X-1}{X+2}+\frac{X}{X+3}-\frac{2X^2+X+1}{X^2+5X+6}=\frac{-9}{2}\)
\(1,\)\(\frac{x+2}{x+3}+\frac{x-1}{x+1}=\frac{2}{x^2+4x+3}+1\)
\(\Rightarrow\frac{\left(x+2\right)\left(x+1\right)}{\left(x+1\right)\left(x+3\right)}+\frac{\left(x-1\right)\left(x+3\right)}{\left(x+1\right)\left(x+3\right)}=\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{\left(x+1\right)\left(x+3\right)}{\left(x+1\right)\left(x+3\right)}\)
\(\Rightarrow\)\(x^2+3x+2+x^2-2x-3=2+x^2+4x+3\)
\(\Rightarrow x^2-3x-6=0\)
.....
\(\frac{x+1}{x-2}+\frac{2x-1}{x-1}=\frac{2}{x^2-3x+2}+\frac{11}{2}\)
\(\Rightarrow\frac{2\left(x+1\right)\left(x-1\right)}{2\left(x-2\right)\left(x-1\right)}+\frac{2\left(2x-1\right)\left(x-2\right)}{2\left(x-1\right)\left(x-2\right)}\)\(=\frac{4}{2\left(x-1\right)\left(x-2\right)}+\frac{22\left(x-1\right)\left(x-2\right)}{2\left(x-1\right)\left(x-2\right)}\)
\(\Rightarrow2x^2-2+4x^2-10x+4=4+22x^2-66x+44\)
.....
\(3,\)\(-2x^2+4x+3\)
\(=-2\left(x^2-2x-\frac{3}{2}\right)\)
\(=-2\left[\left(x^2-2x+1\right)-\frac{5}{2}\right]\)
\(=-2\left(x-1\right)^2+5\)
Đa thức này lớn nhất =5 khi và chỉ khi \(\left(x-1\right)^2\)nhỏ nhất
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
Tìm x
a, -2/3 . x + 4 = 12
b, -3/4 + 1/4 : x = -3
c, ( 2x - 1/3 ) ( 0.5x +0,25 ) = 0
\(a,\frac{-2}{3}x=8\)<=> \(x=-12\)
\(b,\frac{1}{4}:x=-3+\frac{3}{4}\)<=>\(\frac{1}{4}:x=\frac{-9}{4}\)<=>\(x=-9\)
\(c,\orbr{\begin{cases}2x-\frac{1}{3}=0\\0.5x+0.25=0\end{cases}}\)<=>\(\orbr{\begin{cases}x=\frac{1}{6}\\x=\frac{-3}{4}\end{cases}}\)
a) \(\frac{-2}{3}x+4=12\)
\(\Rightarrow\frac{-2}{3}x=12-4\)
\(\Rightarrow\frac{-2}{3}x=8\)
\(\Rightarrow x=8:\frac{-2}{3}\)
\(\Rightarrow x=-12\)
Vậy x = -12
b) \(\frac{-3}{4}+\frac{1}{4}:x=-3\)
\(\Rightarrow\frac{1}{4}:x=-3-\left(\frac{3}{4}\right)\)
\(\Rightarrow\frac{1}{4}:x=\frac{-9}{4}\)
\(\Rightarrow x=\frac{1}{4}:\frac{-9}{4}\)
\(\Rightarrow x=\frac{-1}{9}\)
Vậy \(x=\frac{-1}{9}\)
c) \(\left(2x-\frac{1}{3}\right)\left(0,5x+0,25\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{3}=0\\0,5x+0,25=0\end{cases}}\Rightarrow\orbr{\begin{cases}2x=\frac{1}{3}\\0,5x=-0,25\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{6}\\x=-0,5\end{cases}}\)
Vậy \(x=\frac{1}{6}\)hoặc \(x=-0,5\)
_Chúc bạn học tốt_