giai pt
(x+2)(3x+1)+x^2=4
VH
Những câu hỏi liên quan
giai pt \(x^2-3x=2\sqrt{x-1}-4\)
bài 1 giai cac pt sau
a 11-2x =x-1
b 5(3x+2)=4x+1
c x mũ 2 -4-(x-2)(x-5)
a,\(11-2x=x-1\Leftrightarrow-2x-x=-1-11\Leftrightarrow-3x=-12\Leftrightarrow x=-4\)
b,\(\text{5(3x+2)=4x+1}\Leftrightarrow15x+10=4x+1\Leftrightarrow15x-4x=1-10\Leftrightarrow11x=-9\Leftrightarrow x=\dfrac{-9}{11}\)
c,\(x^2-4-\left(x-2\right)\left(x-5\right)\Leftrightarrow\left(x+2\right)\left(x-2\right)-\left(x-2\right)\left(x-5\right)\Leftrightarrow\left(x-2\right)[\left(x+2\right)-\left(x-5\right)]\Leftrightarrow\left(x-2\right)\left[x+2-x+5\right]\Leftrightarrow\left(x-2\right)7\Leftrightarrow7x-14\)
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C/m pt sau vo nghiem:
x^4-2x^3+3x^2-2x+1=0
Giai pt:
(x^2-4)^2=8x+1
HELP ME
\(x^4-2x^3+3x^2-2x+1=0\)
Chia cả hai vé cho \(x^2\)
\(\Leftrightarrow x^2-2x+3-\dfrac{2}{x}+\dfrac{1}{x^2}\)
\(\Leftrightarrow x^2+2+\dfrac{1}{x^2}-2\left(x+\dfrac{1}{x}\right)+1=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)^2-2\left(x+\dfrac{1}{x}\right)+1=0\)
Đặt x+1/x = a, ta có:
\(a^2-2a+1=0\)
\(\Leftrightarrow\left(a-1\right)^2=0\)
\(\Leftrightarrow a=1\)
\(\Leftrightarrow x+\dfrac{1}{x}=1\)
\(\Leftrightarrow x^2+1=x\)
\(\Leftrightarrow x^2-x+1=0\)
\(\Leftrightarrow x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\)
Do \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+3>0\)
Do đó phương trình vô nghiệm
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giai pt : a. x^4/2x^2+1 + 2x^2+1/x^4=2
b.(x/x-1)^2+(x/x+1)^2=10/9
c. x^3+3x^2-10x-24=0
giai pt : (6x+7)2 (3x+4)(x+1)=6
(6x+7)2.2.(3x+4).6.(x+1) = 72
=> (6x+7)2. (6x+8).(6x+6)= 72
=> (6x+7)2. (6x+7 + 1)(6x+7 - 1) = 72
=> (6x+7)2. [(6x+7)2 - 1] = 72
=> (6x+7)4 - (6x+7)2 = 72 => (6x+7)4 -9.(6x+7)2 + 8.(6x+7)2 - 72 = 0
=> (6x+7)2. [(6x+7)2 - 9] + 8.[(6x+7)2 - 9] = 0
=> [(6x+7)2 + 8].[(6x+7)2 - 9] = 0
=> (6x+7)2 - 9 = 0 Vì (6x+7)2 + 8 > o với mọi x
=> (6x+7)2 = 9 => 6x + 7 = 3 hoặc -3
6x+ 7 =3 => x = -2/3
6x+7 = -3 => x = -5/3
Vậy x = -2/3; -5/3
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(6x +7)2(3x +4)(x +1) =6 <=> (6x +7)2(6x +8)(x +1) = 12
Đặt 6x +7 =t => 6x + 8 = t +1 ; x =(t - 7)/6 ; x +1 = (t -1)/6
Pt trở thành : \(t^2\left(t+1\right)\frac{t-1}{6}=12\Leftrightarrow t^4-t^2-72=0\Leftrightarrow\left(t^2-9\right)\left(t^2+8\right)=0\)
<=> \(t^2-9=0\)( vì t2 +8 >0) <=> t = 3 hay t = -3
t =3 => 6x +7 = 3 => x = -2/3
t= -3 => 6x +7 = -3 => x = -5/3
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giai pt: a) x2+(x2+4)-3x(x2+1)+1=0
cac ban viet buoc giai cho nhe!! minh dang can gap lam !!
1. Cho pt: x2 -2(m+1)x+m20 (1). Tìm m để pt có 2 nghiệm x1 ; x2 thỏa mãn (x1-m)2 + x2m+2.
2. Giai pt: left(x-1right)sqrt{2left(x^2+4right)}x^2-x-2
3. Giai hệ pt: left{{}begin{matrix}frac{1}{sqrt[]{x}}-frac{sqrt{x}}{y}x^2+xy-2y^2left(1right)left(sqrt{x+3}-sqrt{y}right)left(1+sqrt{x^2+3x}right)3left(2right)end{matrix}right.
4. Giai pt trên tập số nguyên x^{2015}sqrt{yleft(y+1right)left(y+2right)left(y+3right)}+1
Đọc tiếp
1. Cho pt: x2 -2(m+1)x+m2=0 (1). Tìm m để pt có 2 nghiệm x1 ; x2 thỏa mãn (x1-m)2 + x2=m+2.
2. Giai pt: \(\left(x-1\right)\sqrt{2\left(x^2+4\right)}=x^2-x-2\)
3. Giai hệ pt: \(\left\{{}\begin{matrix}\frac{1}{\sqrt[]{x}}-\frac{\sqrt{x}}{y}=x^2+xy-2y^2\left(1\right)\\\left(\sqrt{x+3}-\sqrt{y}\right)\left(1+\sqrt{x^2+3x}\right)=3\left(2\right)\end{matrix}\right.\)
4. Giai pt trên tập số nguyên \(x^{2015}=\sqrt{y\left(y+1\right)\left(y+2\right)\left(y+3\right)}+1\)
Giai pt sau:
1/x^2-3x+2 +1/x^2-5x+6 +1/x^2-7x+12 =2(Tất cả =2 nhé!)
=>\(\dfrac{-1}{x-1}+\dfrac{1}{x-2}-\dfrac{1}{x-2}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-4}=2\)
=>\(\dfrac{1}{x-4}-\dfrac{1}{x-1}=2\)
=>\(\dfrac{x-1-x+4}{x^2-5x+4}=2\)
=>2x^2-10x+8=3
=>2x^2-10x+5=0
=>\(x=\dfrac{5\pm\sqrt{15}}{2}\)
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a) Giai PT : 3x - 1 +\(\frac{x-1}{4x}=\sqrt{3x+1}\)
b) Giai hệ PT sau :
\(\left\{{}\begin{matrix}x^3-y^3=4x+2y\\x^2-1=3\left(1-y^2\right)\end{matrix}\right.\)
Câu 1: ĐKXĐ: ...
\(\Leftrightarrow4x\left(3x-1\right)+x-1=4x\sqrt{3x+1}\)
\(\Leftrightarrow12x^2-3x-1-4x\sqrt{3x+1}=0\)
\(\Leftrightarrow16x^2-\left(4x^2+4x\sqrt{3x+1}+3x+1\right)=0\)
\(\Leftrightarrow16x^2-\left(2x+\sqrt{3x+1}\right)^2=0\)
\(\Leftrightarrow\left(2x-\sqrt{3x+1}\right)\left(6x+\sqrt{3x+1}\right)=0\)
\(\Leftrightarrow...\)
Câu 2:
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(x^2-4\right)=y^3+2y\\x^2-4=-3y^2\end{matrix}\right.\)
\(\Leftrightarrow x\left(-3y^2\right)=y^3+2y\)
\(\Leftrightarrow y\left(y^2+3xy+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=0\Rightarrow...\\y^2+3xy+2=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow3xy=-y^2-2\Rightarrow x=\frac{-y^2-2}{3y}\)
\(\Rightarrow\left(\frac{y^2+2}{3y}\right)^2-1=3\left(1-y^2\right)\)
\(\Leftrightarrow\left(\frac{y^2-3y+2}{3y}\right)\left(\frac{y^2+3y+2}{3y}\right)=3\left(1-y^2\right)\)
\(\Leftrightarrow\frac{\left(y-1\right)\left(y-2\right)\left(y+1\right)\left(y+2\right)}{9y^2}=3\left(1-y^2\right)\)
\(\Leftrightarrow\frac{\left(y^2-1\right)\left(y^2-4\right)}{9y^2}=3\left(1-y^2\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}y^2-1=0\\\frac{y^2-4}{9y^2}=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}y^2-1=0\\28y^2=4\end{matrix}\right.\)
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\(3x-1+\frac{x-1}{4x}=\sqrt{3x+1}\)
\(\Leftrightarrow\frac{4x\left(3x-1\right)+x-1}{4x}=\sqrt{3x+1}\)
\(\Leftrightarrow\frac{12x^2-4x+x-1}{4x}=\sqrt{3x+1}\)
\(\Leftrightarrow\frac{12x^2-3x-1}{4x}=\sqrt{3x+1}\)
\(\Leftrightarrow\frac{\left(12x^2-3x-1\right)^2}{16x^2}=3x+1\)
\(\Leftrightarrow\left(12x^2-3x-1\right)^2=16x^2\left(3x+1\right)\)
\(\Leftrightarrow144x^4-120x^3-31x^2+6x+1=0\)
\(\Leftrightarrow144x^4-144x^3+24x^3-24x^2-7x^2+7x-x+1=0\)
\(\Leftrightarrow144x^3\left(x-1\right)+24x^2\left(x-1\right)+7x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(144x^3+24x^2+7x-1\right)=0\)
Tìm được mỗi nghiệm thôi à :v
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