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10 tháng 7 2022
1:
\(\Leftrightarrow\left(x^2+5x+6\right)\left(x^2+5x+4\right)=24\)
\(\Leftrightarrow\left(x^2+5x\right)^2+10\left(x^2+5x\right)=0\)
\(\Leftrightarrow x^2+5x=0\)
=>x=0 hoặc x=-5
3: \(\Leftrightarrow\left(x^2+x+6\right)\left(x^2+x-2\right)=0\)
=>(x+2)(x-1)=0
=>x=-2 hoặc x=1
T
8 tháng 3 2018
Đặt \(x=y-3\).
\(\Rightarrow\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)=\left(y-2\right)\left(y-1\right)\left(y+2\right)\left(y+1\right)=\left(y^2-1\right)\left(y^2-4\right)=40\)
\(\Rightarrow y^2=9\)
\(\Rightarrow x=\hept{\begin{cases}0\\-6\end{cases}}\)
24 tháng 3 2020
(x + 1)(x + 2)(x + 4)(x + 5) = 40
<=> (x + 1)(x + 5)(x + 2)(x + 4) - 40 = 0
<=> (x2 + 6x + 5)(x2 + 6x + 8) - 40 = 0
Đặt x2 + 6x + 5 = a <=> a(a + 3) - 40 = 0
<=> a2 + 3a - 40 = 0
<=> a2 + 8a - 5a - 40 = 0
<=> (a + 8)(a - 5) = 0
<=> \(\orbr{\begin{cases}a+8=0\\a-5=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x^2+6x+5+8=0\\x^2+6x+5-5=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x^2+6x+9+4=0\\x^2+6x=0\end{cases}}\)
<=> \(\orbr{\begin{cases}\left(x+3\right)^2+4=0\left(vn\right)\\x\left(x+6\right)=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=0\\x+6=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=0\\x=-6\end{cases}}\) Vậy S = {0; -6}
BK
4
ND
24 tháng 3 2018
1)
\(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)=40\)
\(\Leftrightarrow\left(x+1\right)\left(x+5\right).\left(x+2\right)\left(x+4\right)-40=0\)
\(\Leftrightarrow\left(x^2+6x+5\right).\left(x^2+6x+8\right)-40=0\)
Đặt \(a=x^2+6x+6\) ta có:
\(\Leftrightarrow\left(a-1\right)\left(a+2\right)-40=0\)
\(\Leftrightarrow a^2+a-2-40=0\)
\(\Leftrightarrow a^2-6x+7x-42=0\)
\(\Leftrightarrow a\left(a-6\right)+7\left(a-6\right)=0\)
\(\Leftrightarrow\left(a-6\right)\left(a+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=6\\a=-7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+6x+6=6\\x^2+6x+6=-7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+6x=0\\x^2+6x+13=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=0\end{matrix}\right.\)
(\(x^2+6x+13=\left(x+3\right)^2+4>0\left(loại\right)\))
Vậy.................
ND
24 tháng 3 2018
3)
\(\left|x+4\right|=\left|3-2x\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=3-2x\\x+4=-3+2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\-x+7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=7\end{matrix}\right.\)
Vậy..........
PN
1
8 tháng 9 2016
E=x5-5x4+5x3-5x2+5x-1
=x5-4x4+x3-4x2+x-x4+4x3-x2+4x-1
=x(x4-4x3+x2-4x+1)-(x4-4x3+x2-4x+1)
=(x-1)(x4-4x3+x2-4x+1)
Tại x=4 ta có:
E=(4-1)(44-4*43+42-4*4+1)
=3*(256-256+16-16+1)
=3*1
=3
DA
2
TG
16 tháng 1 2021
a) \(x^2+2x=\left(x-2\right).3x\)
\(\Leftrightarrow x^2+2x=3x^2-6x\)
\(\Leftrightarrow x^2+2x-3x^2+6x=0\)
\(\Leftrightarrow-2x^2+8x=0\)
\(\Leftrightarrow-2x\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x=0\\x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy S = {0;4}
b) \(x^3+x^2-x-1=0\)
\(\Leftrightarrow x^2\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x^2-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\mp1\end{matrix}\right.\)
Vậy: S = {-1; 1}
c) \(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)=40\)
\(\Leftrightarrow\left[\left(x+1\right)\left(x+5\right)\right]\left[\left(x+2\right)\left(x+4\right)\right]=40\)
\(\Leftrightarrow\left(x^2+5x+x+5\right)\left(x^2+4x+2x+8\right)=40\)
\(\Leftrightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)=40\)
Đặt x2 + 6x + 5 = t
\(\Leftrightarrow t.\left(t+3\right)=40\)
\(\Leftrightarrow t^2+3t=40\)
\(\Leftrightarrow t^2+2.t.\dfrac{3}{2}+\dfrac{9}{4}=\dfrac{169}{4}\)
\(\Leftrightarrow\left(t+\dfrac{3}{2}\right)^2=\dfrac{169}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}t+\dfrac{3}{2}=\dfrac{13}{2}\\t+\dfrac{3}{2}=-\dfrac{13}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{13}{2}-\dfrac{3}{2}=\dfrac{10}{2}=5\\t=-\dfrac{13}{2}-\dfrac{3}{2}=-\dfrac{16}{2}=-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+6x+5=5\\x^2+6x+5=-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+6x=0\\x^2+6x+13=0\end{matrix}\right.\)
Mà: \(x^2+6x+13=x^2+2.x.3+9+4=\left(x+3\right)^2+4\ne0\)
=> x2 + 6x = 0
<=> x. (x + 6) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)
Vậy S = {0; -6}
16 tháng 1 2021
a) Ta có: \(x^2+2x=\left(x-2\right)\cdot3x\)
\(\Leftrightarrow x\left(x+2\right)-3x\left(x-2\right)=0\)
\(\Leftrightarrow x\left[\left(x+2\right)-3\left(x-2\right)\right]=0\)
\(\Leftrightarrow x\left(x+2-3x+6\right)=0\)
\(\Leftrightarrow x\left(-2x+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-2x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\-2x=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy: S={0;4}
b) Ta có: \(x^3+x^2-x-1=0\)
\(\Leftrightarrow x^2\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\cdot\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\cdot\left(x-1\right)\cdot\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2\cdot\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+1\right)^2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
Vậy: S={-1;1}
c) Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)=40\)
\(\Leftrightarrow\left(x+1\right)\left(x+5\right)\left(x+2\right)\left(x+4\right)-40=0\)
\(\Leftrightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)-40=0\)
\(\Leftrightarrow\left(x^2+6x\right)^2+13\left(x^2+6x\right)+40-40=0\)
\(\Leftrightarrow\left(x^2+6x\right)^2+13\left(x^2+6x\right)=0\)
\(\Leftrightarrow\left(x^2+6x\right)\left(x^2+6x+13\right)=0\)
\(\Leftrightarrow x\left(x+6\right)\left(x^2+6x+13\right)=0\)
mà \(x^2+6x+13>0\forall x\)
nên \(x\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)
Vậy: S={0;-6}
BV
2
3 tháng 2 2017
2, đặt x2+x=a ta có:
a+4a-12=0\(\Leftrightarrow\)( a+2.2a+4)-16=0 \(\Leftrightarrow\) (a+2)2-42=0 \(\Leftrightarrow\)(a-2)(a+6)=0
\(\left[\begin{matrix}a-2=0\\a+6+0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}a=2\\a=-6\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x^2+x=2\\x^2+x=-6\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[\begin{matrix}x^2+x-2=0\\x^2+x+6=0\left(vl\right)\end{matrix}\right.\)
\(\Leftrightarrow\)x2-x+2x-2=0\(\Leftrightarrow\)x(x-1)+2(x-1)=0\(\Leftrightarrow\left[\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
vậy pt có tập nghiệm là S=\(\left\{-2;1\right\}\)
3, (x+1) (x+2) (x+4) (x+5)= 40
\(\Leftrightarrow\)(x+1)(x+5)(x+2)(x+4)=40
\(\Leftrightarrow\)(x2+6x+5)(x2+6x+8)-40=0
đặt x2+6x+5=y ta có
y(y+3)-40=0\(\Leftrightarrow\)y2+2.\(\frac{3}{2}y\)+\(\frac{9}{4}\)-\(\frac{169}{4}\)=0\(\Leftrightarrow\)(y+\(\frac{3}{2}\))2-(\(\frac{13}{2}\))2=0\(\Leftrightarrow\)(y-5)(y+8)=0\(\Leftrightarrow\left[\begin{matrix}y-5=0\\y+8=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[\begin{matrix}y=5\\y=-8\end{matrix}\right.\)
\(\Leftrightarrow\left[\begin{matrix}x^2+6x+5=5\\x^2+6x+5=-8\end{matrix}\right.\)\(\Leftrightarrow\left[\begin{matrix}x^2+6x=0\\x^2+6x+13=0\left(vl\right)\end{matrix}\right.\)\(\Leftrightarrow\)x(x+6)=0\(\Leftrightarrow\left[\begin{matrix}x=0\\x=-6\end{matrix}\right.\)
vậy pt có tập nghiêm là S=\(\left\{-6;0\right\}\)
2 tháng 2 2017
2) (x2 +x )+4 (x2 +x) -12= 0
đặt x2+x=a rồi thay vào , biến đổi thành HDT bình phương là đc
3) (x+1) (x+2) (x+4) (x+5)= 40
nhân (x+1)(x+5)và (x+2)(x+4)rồi đặt biến phụ rồi làm giống câu trên (chuyển 40 sang vế phải)
TN
2
28 tháng 11 2016
mik chẳng hỉu gì cả
@@#$^^#^&%&$&%$##%$#@##@$#@#$%^*%^&^%$%
15 tháng 3 2021
Bài 1:
c) ĐKXĐ: \(x\notin\left\{\dfrac{1}{4};-\dfrac{1}{4}\right\}\)
Ta có: \(\dfrac{3}{1-4x}=\dfrac{2}{4x+1}-\dfrac{8+6x}{16x^2-1}\)
\(\Leftrightarrow\dfrac{-3\left(4x+1\right)}{\left(4x-1\right)\left(4x+1\right)}=\dfrac{2\left(4x-1\right)}{\left(4x+1\right)\left(4x-1\right)}-\dfrac{6x+8}{\left(4x-1\right)\left(4x+1\right)}\)
Suy ra: \(-12x-3=8x-2-6x-8\)
\(\Leftrightarrow-12x-3-2x+10=0\)
\(\Leftrightarrow-14x+7=0\)
\(\Leftrightarrow-14x=-7\)
\(\Leftrightarrow x=\dfrac{1}{2}\)(nhận)
Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)
Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)=40\)
\(\Leftrightarrow\left(x+1\right)\left(x+5\right)\left(x+2\right)\left(x+4\right)-40=0\)
\(\Leftrightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)-40=0\)
Đặt \(y=x^2+6x+5\), ta có:
\(y\left(y+3\right)-40=0\)
\(\Leftrightarrow\left(y-5\right)\left(y+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}y=5\\y=-8\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-6\end{cases}}\)