Giải phương trình
1/ (x + 1)(x + 2) + 1/ (x + 2)(x + 3) +1/ (x +3)(x + 4) = 1/10
giải chi tiết gium nha
Giải phương trình
1) 7 - (2x + 4) = - (x +4)
2) ( x - 1) - (2x -1) = 9 - x
3) (x + 1) - (2x - 3) = ( 2x - 1)( x + 5)
\(1,\Leftrightarrow7-2x-4=-x-4\)
\(\Leftrightarrow x-2x=-4-7+4\)
\(\Leftrightarrow-x=-7\)
\(\Leftrightarrow x=7\)
Vậy \(S=\left\{7\right\}\)
\(2,\Leftrightarrow x-1-2x+1=9-x\)
\(\Leftrightarrow x+x-2x=9-1+1\)
\(\Leftrightarrow0x=9\)
\(\Rightarrow x\in\varnothing\)
Vậy \(S=\left\{\varnothing\right\}\)
\(3,\Leftrightarrow2x^2+3x-2x+3=2x^2+10x-x-5\)
\(\Leftrightarrow2x^2-2x^2+3x-2x-10x+x=-5-3\)
\(-8x=-8\)
\(\Rightarrow x=1\)
Vậy \(S=\left\{1\right\}\)
3/ TÍNH GIÁ TRỊBIỂU THỨC
a)(-3x2)3+ 4x–9 –27x6tại x = 2
b)2x3(x –8)+ x4(x + 7) –(x5+ 9x4–16x3+ x2+ x –1 )tại x = 10
Giải ra chi tiết giùm mình nha
a: Ta có: \(-\left(-3x^2\right)^3+4x-9-27x^6\)
\(=27x^6-27x^6+4x-9\)
=4x-9
=-1
giải phương trình
1)\(\sqrt{2x+5}+\sqrt{x-1}=8\)
2)\(\sqrt{1-x}+\sqrt{4+x}=3\)
giải phương trình
x = 1/2 . x + 1/4 . x + 1/7 .x + 3
dấu . là đấu nhân nha giải chi tiết giùm nha
\(x=\frac{1}{2}x+\frac{1}{4}x+\frac{1}{7}x+3\)
\(\Rightarrow x=x\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{7}\right)+3\)
\(\Rightarrow x=\frac{25}{28}x+3\Rightarrow x-\frac{25}{28}x=3\Rightarrow x\left(1-\frac{25}{28}\right)=3\Rightarrow x.\frac{3}{28}=3\Rightarrow x=28\)
Vậy x = 28
Đặt x làm nhân tử chung : x(1/2+1/4+1/7)+3=x
Qui đồng cái tổng đấy chuyển vế là ra
\(\frac{1}{2}x+\frac{1}{4}x+\frac{1}{7}x+3=x\)
<=>x(1/2+1/4+1/7)+3=x
<=>x(14/28+7/28+4/28)+3=x
<=>x28/28(14/28+7/28+4/28)+84/28=x.28/28
=>x.28(14+7+4)+84=x.28
<=>x.28.25+84=x.28
<=>
giải phương trình
1)\(\sqrt{31-x}=x-1\)
2)\(3\sqrt{x^2-1}=x^2+1\)
3)\(\sqrt{x^2-3x+5}+x=3x+7\)
Giải các phương trình
1,\(x\left(x-1\right)=2\left(x-1\right)\)
2, \(\left(x+2\right)\left(2x-3\right)=x^2-4\)
3, \(x^2+3x+2=0\)
4, \(5x^2+5x+3=0\)
5, \(x^3+x^2-12x=0\)
1, x(x-1)=2(x-1)
<=> x(x-1)-2(x-1)=0
<=> (x-2)(x-1)=0
<=>x=2 hoặc x=1
vậy ...
2, (x+2)(2x-3)=x^2 -4
<=>(x+2)(2x-3)=(x-2)(x+2)
<=> (x+2)(2x-3)-(x-2)(x+2)=0
<=> (x+2)(2x-3-x+2)=0
<=> x=-2 hoặc x=1
vây...
3,x^2 +3x +2=0
<=> x^2 +x+2x+2=0
<=>(x+2)(x+1)=0
<=> x=-2 hoặc x=-1
vậy ...
5, x^3+x^2-12x =0
<=> x(x^2+x-12)=0
<=>x(x^2-3x+4x-12)=0
<=>x(x+4)(x-3)=0
<=> x=0 hoặc x=-4 hoặc x=3
vậy ...
Bài 1: Giải phương trình
1) \(\sqrt{4x^2+12x+9}=2-x\left(vớix\le0\right)\)
2) \(\sqrt{x^4+2x^2+1}=x^2+5x+4\) ( với \(x^2+5x+4>0\))
3) \(\sqrt{5x+1}=4\)
4) \(\sqrt{3-x}=7\)
Câu 2,3,4 nx thôi ạ. Câu 1 có bạn giúp r ạ
1)\(\sqrt{4x^2+12x+9}=2-x\)
\(\Leftrightarrow\sqrt{\left(2x+3\right)^2}=2-x\)
\(\Leftrightarrow\left|2x+3\right|=2-x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=2-x\\2x+3=x-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-5\end{matrix}\right.\)
\(\)
2)\(\sqrt{x^4+2x^2+1}=x^2+5x+4\) ĐK:\(x\ge-1\)
\(\Leftrightarrow\sqrt{\left(x^2+1\right)^2}=x^2+5x+4\)
\(\Leftrightarrow\left|x^2+1\right|=x^2+5x+4\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+1=x^2+5x+4\\x^2+1=-x^2-5x-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-3\\2x^2+5x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\2\left(x+\dfrac{5}{4}\right)^2+\dfrac{15}{8}=0\left(voli\right)\end{matrix}\right.\)
giải phương trình
1)\(2\left(x-3\right)+1=2\left(x+1\right)-9\)
2)\(\dfrac{5-x}{2}=\dfrac{3x-4}{6}\)
3) \(\left(x-1\right)^2+\left(x+2\right)\left(x-2\right)=\left(2x+1\right)\left(x-3\right)\)
4)\(\left(x+5\right)\left(x-1\right)-\left(x+1\right)\left(x+2\right)=1\)
5) \(\dfrac{6x-1}{15}-\dfrac{x}{5}=\dfrac{2x}{3}\)
6)\(\dfrac{5\left(x-2\right)}{2}-\dfrac{x+5}{3}=1-\dfrac{4\left(x-3\right)}{5}\)
\(1,2\left(x-3\right)+1=2\left(x+1\right)-9\\ \Rightarrow2x-6+1=2x+2-9\\ \Rightarrow2x-5=2x-7\\ \Rightarrow-2=0\left(vô.lí\right)\)
\(2,\dfrac{5-x}{2}=\dfrac{3x-4}{6}\\ \Rightarrow30-6x=6x-8\\ \Rightarrow12x=38\\ \Rightarrow x=\dfrac{19}{6}\)
\(3,\left(x-1\right)^2+\left(x+2\right)\left(x-2\right)=\left(2x+1\right)\left(x-3\right)\\ \Rightarrow x^2-2x+1+x^2-4=2x^2-6x+x-3\\ \Rightarrow2x^2-2x-3=2x^2-5x-3\\ \Rightarrow3x=0\\ \Rightarrow x=0\)
\(4,\left(x+5\right)\left(x-1\right)-\left(x+1\right)\left(x+2\right)=1\\ \Rightarrow x^2+5x-x-5-x^2-2x-x-2=1\\ \\ \Rightarrow x-7=1\\ \Rightarrow x=8\)
\(5,\dfrac{6x-1}{15}-\dfrac{x}{5}=\dfrac{2x}{3}\\ \Rightarrow\dfrac{6x-1}{15}-\dfrac{3x}{15}=\dfrac{10x}{15}\\ \Rightarrow6x-1-3x=10x\\ \Rightarrow3x-1=10x\\ \Rightarrow7x=-1\\ \Rightarrow x=\dfrac{-1}{7}\)
\(6,\dfrac{5\left(x-2\right)}{2}-\dfrac{x+5}{3}=1-\dfrac{4\left(x-3\right)}{5}\\ \Rightarrow\dfrac{75\left(x-2\right)}{30}-\dfrac{10\left(x+5\right)}{30}=\dfrac{30}{30}-\dfrac{24\left(x-3\right)}{30}\\ \Rightarrow75\left(x-2\right)-10\left(x+5\right)=30-24\left(x-3\right)\\ \Rightarrow75x-150-10x-50=30-24x+72\\ \Rightarrow65x-200=102-24x\\ \Rightarrow89x=302\\ \Rightarrow x=\dfrac{320}{89}\)
giải phương trình
a, (x + 3)^4 + (x + 5)^4 = 16
b, (x - 2)^4 + (x - 3)^4 = 1
giải chi tiết giùm nha
a/ (x + 3)4 + (x + 5)4 = 16
=> (x2 + 6x + 9)2 + (x2 + 10x + 25)2 = 16
=> x4 + 36x2 + 81 + 12x3 + 108x + 18x2 + x4 + 100x2 + 625 + 20x3 + 500x + 50x2 = 16
=> 2x4 + 32x3 + 204x2 + 608x + 690 = 0
=> 2(x + 3)(x + 5)(x2 + 8x + 23) = 0
=> (x + 3)(x + 5)(x2 + 8x + 23) = 0
=> x = -3
hoặc x = -5
hoặc x2 + 8x + 23 = 0 , mà x2 + 8x + 23 > 0 => pt vô nghiệm
Vậy x = -3 , x = -5
\(a.\) \(\left(x+3\right)^4+\left(x+5\right)^4=16\) \(\left(1\right)\)
Đặt \(y=x+4\), khi đó, phương trình \(\left(1\right)\) trở thành:
\(\left(y-1\right)^4+\left(y+1\right)^4=16\)
\(\Leftrightarrow\) \(y^4-4y^3+6y^2-14y+1+y^4+4y^3+6y^2+14y+1=16\)
\(\Leftrightarrow\) \(2y^4+12y^2+2=16\)
\(\Leftrightarrow\) \(y^4+6y^2+1=8\)
\(\Leftrightarrow\) \(y^4+6y^2-7=0\)
\(\Leftrightarrow\) \(\left(y^2-1\right)\left(y^2+7\right)=0\) \(\left(1'\right)\)
Vì \(y^2+7>0\) với mọi \(y\) (vì \(y^2\ge0\) ) nên từ \(\left(1'\right)\), suy ra \(y^2-1=0\), hay \(y^2=1\) \(\Leftrightarrow\) \(^{y=1}_{y=-1}\)
Do đó, ta tìm được \(x_1=-3\) hoặc \(x_2=-5\)
Vậy, \(S=\left\{-3;-5\right\}\)