GTTT của 4-3x cộng với 1=2x
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a,\(2x+1=0< =>2x=-1< =>x=-\frac{1}{2}\)
b,\(\left(x+1\right)\left(2x-1\right)=0< =>\orbr{\begin{cases}x+1=0\\2x-1=0\end{cases}< =>\orbr{\begin{cases}x=-1\\x=\frac{1}{2}\end{cases}}}\)
c,\(1-4x^2=0< =>\left(1-2x\right)\left(1+2x\right)=0< =>\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)
d,\(2x^2-3x=0< =>x\left(2x-3\right)=0< =>\orbr{\begin{cases}x=0\\x=\frac{3}{2}\end{cases}}\)
\(\dfrac{x+10}{x-2}+\dfrac{x-18}{x+2}+\dfrac{x+2}{x^2-4}=\dfrac{\left(x+10\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-18\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2+12x+20+x^2-16x-36+x+2}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x^2-3x-14}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(2x^2+4x\right)-\left(7x+14\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x\left(x+2\right)-7\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(2x-7\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x-7}{x-2}\)
\(\sqrt{3x^2+6x+12}+\sqrt{5x^4-10x^2+9}\\ =\sqrt{3\left(x^2+2x+1\right)+9}+\sqrt{5\left(\left(x^2\right)^2-2x^2+1\right)+4}\\ =\sqrt{3\left(x+1\right)^2+9}+\sqrt{5\left(x^2-1\right)^2+4}\)
do: \(+\left(x+1\right)^2\ge0\Rightarrow3.\left(x+1\right)^2+9\ge9\Rightarrow\sqrt{3\left(x+1\right)^2+9}\ge\sqrt{9}=3\)(1)\(+\left(x^2-1\right)^2\ge0\Rightarrow5\left(x^2-1\right)^2+4\ge4\Rightarrow\sqrt{5\left(x^2-1\right)^2+4}\ge\sqrt{4}=2\)(2)
từ (1) và(2)\(\Rightarrow\sqrt{3\left(x+1\right)^2+9}+\sqrt{5\left(x^2-1\right)^2+4}\ge3+2=5\)
câu b bạn làm tương tự
\(\dfrac{3-3x}{2x}+\dfrac{3x-1}{2x-1}+\dfrac{11x-5}{2x-4x^2}\\ =\dfrac{\left(3-3x\right)\left(1-2x\right)}{2x\left(1-2x\right)}-\dfrac{2x\left(3x-1\right)}{2x\left(1-2x\right)}+\dfrac{11x-5}{2x\left(1-2x\right)}\\ =\dfrac{3-9x+6x^2}{2x\left(1-2x\right)}-\dfrac{6x^2-2x}{2x\left(1-2x\right)}+\dfrac{11x-5}{2x\left(1-2x\right)}\\ =\dfrac{3-9x+6x^2-6x^2+2x+11x-5}{2x\left(1-2x\right)}\\ =\dfrac{-2}{2x\left(1-2x\right)}\\ =\dfrac{-1}{x\left(1-2x\right)}\)
2\(x\) - 3\(x\) = 3
-\(x\) = 3
\(x\) = - 3
Thay \(x\) = - 3 vào biểu thức 5\(x\) - y = 4 ta có:
5.(-3) - y = 4
-15 - y = 4
-15 - 4 = y
y = - 19
Vậy (\(x;y\)) = (-3; -19)
a: 3x-2=2x-3
=>x=-1
b: 2x+3=5x+9
=>-3x=6
=>x=-2
c: 5-2x=7
=>2x=-2
=>x=-2
d: 10x+3-5x=4x+12
=>5x+3=4x+12
=>x=9
e: 11x+42-2x=100-9x-22
=>9x+42=78-9x
=>18x=36
=>x=2
f: 2x-(3-5x)=4(x+3)
=>2x-3+5x=4x+12
=>7x-3=4x+12
=>3x=15
=>x=5
a: \(=\dfrac{2x^4-2x^2-3x^3+3x+6x^2-6-3x+7}{x^2-1}=2x^2-3x+6+\dfrac{-3x+7}{x^2-1}\)
Để số dư là 0 thì -3x+7=0
hay x=7/3
b: \(=\dfrac{x^5+x^3+2x^4+2x^2+2x^3+2x-2x^2-2-x-1}{x^2+1}\)
\(=x^3+2x^2+2x-2+\dfrac{-x-1}{x^2+1}\)
Để số dư là 0 thì -x-1=0
hay x=-1