3x+3^x+2=180
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126-2.(x-1)=20 120+3.(x-3)=180
2.(x-1)=126-20 3.(x-3)=180-120
2.(x-1)=106 3.(x-3)=60
x-1=106:2 x-3=60:3
x-1=53 x-3=20
x=53+1 x=20+3
x=54 x=23
\(\dfrac{20x^2+120x+180}{\left(3x+5\right)^2-4x^2}+\dfrac{5x^2-25}{9x^2-\left(2x+5\right)^2}-\dfrac{\left(2x+3\right)^2-x^2}{3\left(x^2+8x+15\right)}\)
\(=\dfrac{20\left(x^2+6x+9\right)}{\left(3x+5+2x\right)\left(3x+5-2x\right)}+\dfrac{5\left(x-5\right)\left(x+5\right)}{\left(3x-2x-5\right)\left(3x+2x+5\right)}-\dfrac{\left(2x+3-x\right)\left(2x+3+x\right)}{3\left(x+3\right)\left(x+5\right)}\)
\(=\dfrac{20\left(x+3\right)^2}{5\left(x+1\right)\cdot\left(x+5\right)}+\dfrac{5\left(x-5\right)\left(x+5\right)}{5\left(x+1\right)\left(x-5\right)}-\dfrac{\left(x+3\right)\cdot3\left(x+1\right)}{3\left(x+3\right)\left(x+5\right)}\)
\(=\dfrac{4\left(x+3\right)^2}{\left(x+1\right)\left(x+5\right)}+\dfrac{x+5}{x+1}-\dfrac{x+1}{x+5}\)
\(=\dfrac{4\left(x+3\right)^2+\left(x+5\right)^2-\left(x+1\right)^2}{\left(x+1\right)\left(x+5\right)}\)
\(=\dfrac{4x^2+24x+36+x^2+10x+25-x^2-2x-1}{\left(x+1\right)\cdot\left(x+5\right)}\)
\(=\dfrac{4x^2+32x+60}{\left(x+1\right)\left(x+5\right)}=\dfrac{4\left(x^2+8x+15\right)}{\left(x+1\right)\left(x+5\right)}\)
\(=\dfrac{4\left(x+3\right)\cdot\left(x+5\right)}{\left(x+1\right)\left(x+5\right)}=\dfrac{4x+12}{x+1}\)
b) Giải:
Ta có: \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) và x + y + z = 180
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}=\frac{x+y+z}{1+2+3}=\frac{180}{6}=30\)
+) \(\frac{x}{1}=30\Rightarrow x=30\)
+) \(\frac{y}{2}=30\Rightarrow y=60\)
+) \(\frac{z}{3}=30\Rightarrow z=90\)
Vậy bộ số \(\left(x;y;z\right)\) là \(\left(30;60;90\right)\)
c) Sai đề
A) x/2=y/3=z/4 và 180 - 2x -y-z=0
ta có :
180-2x-y-z=0
=> 2x-y-z=180
Theo bài ra ta có :
x/2=y/3=z/4
=> 2x/4=y/3=z/4
Áp dụng t/c của dãy tỷ số bằng nhau ta có :
2x/4=y/3=z/4=2x-y-z/4-3-4=180/-3=-60
=> 2x=-240 => x= -120
y=-180
z=-240
các câu còn lại tự làm đc mà k đc hỏi mk
a.
\(6x^2-3x=7x^2\)
\(\Leftrightarrow-x^2-3x=0\)
\(\Leftrightarrow x^2+3x=0\)
\(\Leftrightarrow x\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
a.
\(6x^2-3x=7x^2\)
\(\Leftrightarrow6x^2-7x^2-3x=0\)
\(\Leftrightarrow-x^2-3x=0\)
\(\Leftrightarrow-x\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
vậy,...
tan(2x+10o)+cot(x)=0
<=> tan(2x+10o)+tan(90o-x)=0
<=>tan(x+100o)*[1-tan(2x-10o)*tan(90o-x)]=0
*tan(x+100o)=0 => x=....
*1-tan(2x-10o)*tan(90o-x)=0
<=> tan(2x-10o)=tanx <=> x=....
Bài 1 :
a) 2x3-3+3x2+8=0
b) x3-1=0
Bài 2 :
a) (x2-5x)2 + 10.(x2-5x)+24=0
b) (x+2)(x+3)(x-5)(x-6)=180
Bài 1:
a) Bạn xem lại đề
b)
\(x^3-1=0\)
\(\Leftrightarrow (x-1)(x^2+x+1)=0\)
Vì \(x^2+x+1=x^2+2.\frac{1}{2}x+(\frac{1}{2})^2+\frac{3}{4}=(x+\frac{1}{2})^2+\frac{3}{4}\geq \frac{3}{4}>0\)
\(\Rightarrow x^2+x+1\neq 0\)
Do đó: \(x-1=0\Rightarrow x=1\) là nghiệm duy nhất
Bài 2:
a) \((x^2-5x)^2+10(x^2-5x)+24=0\)
\(\Leftrightarrow (x^2-5x)^2+2.5(x^2-5x)+5^2-1=0\)
\(\Leftrightarrow (x^2-5x+5)^2-1=0\)
\(\Leftrightarrow (x^2-5x+5-1)(x^2-5x+5+1)=0\)
\(\Leftrightarrow (x^2-5x+4)(x^2-5x+6)=0\)
\(\Leftrightarrow (x-1)(x-4)(x-2)(x-3)=0\)
\(\Rightarrow \left[\begin{matrix} x-1=0\\ x-4=0\\ x-2=0\\ x-3=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=1\\ x=4\\ x=2\\ x=3\end{matrix}\right.\)
b)
\((x+2)(x+3)(x-5)(x-6)=180\)
\(\Leftrightarrow [(x+2)(x-5)][(x+3)(x-6)]=180\)
\(\Leftrightarrow (x^2-3x-10)(x^2-3x-18)=180\)
\(\Leftrightarrow a(a-8)=180\) (đặt \(x^2-3x-10=a\) )
\(\Leftrightarrow a^2-8a+16-196=0\)
\(\Leftrightarrow (a-4)^2-14^2=0\)
\(\Leftrightarrow (a-4-14)(a-4+14)=0\Leftrightarrow (a-18)(a+10)=0\)
\(\Rightarrow a=18\) hoặc $a=-10$
+) Nếu $a=18$ thì \(x^2-3x-10=18\)
\(\Leftrightarrow x^2-3x-28=0\)
\(\Leftrightarrow (x-7)(x+4)=0\Rightarrow \left[\begin{matrix} x=7\\ x=-4\end{matrix}\right.\)
+) Nếu $a=-10$ thì \(x^2-3x-10=-10\Leftrightarrow x^2-3x=0\Leftrightarrow x(x-3)=0\)
\(\Leftrightarrow \left[\begin{matrix} x=0\\ x=3\end{matrix}\right.\)
Vậy pt có 4 nghiệm \(x\in \left\{7;-4;0;3\right\}\)
\(9^2-\left(3x+7\right)=6^2\cdot5\cdot14:180\)
\(81-\left(3x+7\right)=14\)
\(3x+7=81-14\)
\(3x+7=67\)
\(3x=67-7\)
\(3x=60\)
\(x=60:3\)
\(x=20\)
Vậy x = 20
a-26=26
a- =26+26
a =52
52/5=10,4=b/4
b=41,6