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Câu 1 :
a. \(4x-5=23\\ \Leftrightarrow4x=23+5\\ \Leftrightarrow4x=28\\ \Leftrightarrow x=7\)
b.
|-2x|=5x+14
Nếu - 2x > 0 => x < 0 thì |-2x|= - 2x, ta có pt: -2x = 5x+14
<=> - 2x = 5x + 14
<=> - 2x - 5x = 14
<=> - 7x = 14
<=> x = - 2 (thoã mãn)
Nếu - 2x < 0 => x > 0 thì |-2x|= = -(- 2x) = 2x.
Ta có pt: 2x = 5x + 14
<=> - 3x = 14
<=> x = \(-\dfrac{14}{3}\)
Vậy pt có nghiệm x = - 2
c) \(\dfrac{x+1}{x-1}-\dfrac{1}{x+1}=\dfrac{x^2+2}{x^2-1}\\ ĐKXĐ:x\ne1;x\ne-1\\ \Leftrightarrow\dfrac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{1\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2+2}{\left(x-1\right)\left(x+1\right)}\\ \Leftrightarrow x^2+x+x+1-x+1=x^2+2\\ \Leftrightarrow x^2+x+x-x-x^2=2-1-1\\ \Leftrightarrow x=0\left(nhận\right)\)
\(a,4x-5=23\)
\(\Leftrightarrow4x=23+5\)
\(\Leftrightarrow4x=28\)
\(\Leftrightarrow x=7\)
\(b,\left|-2x\right|=5x+14\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=5x+14\\2x=-5x-14\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x-14=0\\7x+14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x=14\\7x=-14\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{14}{3}\\x=-2\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{14}{3};-2\right\}\)
\(c,\Leftrightarrow\dfrac{\left(x+1\right)\left(x+1\right)-x+1-x^2-2}{x^2-1}=0\)
\(\Leftrightarrow x^2+x+x+1-x+1-x^2-2=0\)
\(\Leftrightarrow x=0\)
Vậy \(S=\left\{0\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b ) Ta có : 3x2 - 7x - 6
= 3x2 - 9x + 2x - 6
= 3x (x - 3) + 2(x - 3)
= (x - 3)(3x + 2)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(A=\left(-5x+4\right)\left(3x-2\right)+\left(-2x+3\right)\left(x-2\right)\)
\(=-15x^2+10x+12x-8=-15x^2+22x-8\)
Thay x = -2 vào biểu thức ta có : \(-15\left(-2\right)^2+22\left(-2\right)-8\)
\(=-15.4-44-8=-112\)
b, \(B=\left(x-9\right)\left(2x+3\right)-2\left(x+7\right)\left(x-5\right)\)
\(=2x^2+3x-18x-27=2x^2-15x-27\)
Thay x = -1/2 vào biểu thức ta có : \(2\left(-\frac{1}{2}\right)^2-15\left(-\frac{1}{2}\right)-27\)
\(=2.\frac{1}{4}+\frac{15}{2}-27=\frac{11}{2}+\frac{15}{2}+27=40\)
Bài làm:
a) \(A=\left(-5x+4\right)\left(3x-2\right)+\left(-2x+3\right)\left(x-2\right)\)
\(A=-15x^2+22x-8-2x^2+7x-6\)
\(A=-17x^2+29x-14\)
Thay x = -2 vào ta được:
\(A=-17.\left(-2\right)^2+29.\left(-2\right)-14\)
\(A=-68-58-14\)
\(A=-140\)
b) \(B=\left(x-9\right)\left(2x+3\right)-2\left(x+7\right)\left(x-5\right)\)
\(B=2x^2-15x-27-2\left(x^2+2x-35\right)\)
\(B=2x^2-15x-27-2x^2-4x+70\)
\(B=-19x+43\)
Thay x = -1/2 vào B ta được:
\(B=-19.\left(-\frac{1}{2}\right)+43=\frac{19}{2}+43=\frac{105}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x-1\right)^3-x\left(x-1\right)^2=5x\left(2-x\right)-11\left(x+2\right)\)
\(< =>\left(x-1+x\right)\left(x-1\right)^2=10x-5x^2-11x-22\)
\(< =>-x^2+x-1-10x+5x^2+11x+22=0\)
\(< =>4x^2+3x+21=0\)
\(< =>\left(2x\right)^2+2.2x.\frac{3}{4}+\left(\frac{3}{4}\right)^2+20\frac{9}{25}=0\)
\(< =>\left(2x+\frac{3}{4}\right)^2+20\frac{9}{25}=0\)
Do \(\left(2x+\frac{3}{4}\right)^2\ge0=>\left(2x+\frac{3}{4}\right)^2+20\frac{9}{25}\ge20\frac{9}{25}>0\)
Vậy phương trình vô nghiệm
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Ta có :
\(N=x^2\left(y-1\right)-5x\left(1-y\right)=x^2\left(y-1\right)+5x\left(y-1\right)=x\left(x+5\right)\left(y-1\right)\)
Thay x = -20 ; y = 1001 ta được :
\(-20\left(-20+5\right)\left(1001-1\right)=-20.\left(-15\right).1000=300000\)
b, Ta có : \(x\left(x-y\right)^2-y\left(x-y\right)^2+xy^2-x^2y=\left(x-y\right)^3+xy\left(x-y\right)\)
\(=\left(x-y\right)^4\left(1+xy\right)\)
Thay x - y = 7 ; xy = 9 ta được :
\(7^4.\left(1+9\right)=2401.10=24010\)
N = x2( y - 1 ) - 5x( 1 - y )
= x2( y - 1 ) + 5x( y - 1 )
= x( y - 1 )( x + 5 )
Tại x = -20 ; y = 1001 ta được :
N = -20( 1001 - 1 )( -20 + 5 )
= -20.1000.(-15)
= 1000.300
= 300 000
Q = x( x - y )2 - y( x - y )2 + xy2 - x2y
= x( x - y )2 - y( x - y )2 - xy( x - y )
= ( x - y )[ x( x - y ) - y( x - y ) - xy ]
= ( x - y )( x2 - xy - xy + y2 - xy )
= ( x - y )( x2 - 3xy + y2 )
= ( x - y )[ ( x2 - 2xy + y2 ) + 2xy - 3xy ]
= ( x - y )[ ( x - y )2 - xy ]
= 7[ 72 - 9 ]
= 7( 49 - 9 )
= 7.40 = 280
![](https://rs.olm.vn/images/avt/0.png?1311)
a) x+5x2=0
=>x(1+5x)=0
=>x=0 hoặc 1+5x=0 =>x=\(\dfrac{1}{5}\)
b)x+1=(x+1)2
=>(x+1)-(x+1)2=0
=>(x+1)(1-x-1)=0
=>-x(x+1)=0
=>\(\left[{}\begin{matrix}-x=0\Rightarrow x=0\\x+1=0\Rightarrow x=-1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
câu c,d,e lạ quá nhưng thui you viết thế nào tui làm thế ấy sai đừng trách
a) 5x(x-1)=x-1
<=> 5x(x-1)-(x-1)=0
<=>(x-1)(5x-1)=0
<=>x-1=0 hoặc 5x-1=0
<=>x=1 hoặc \(\frac{1}{5}\)
b) 2(x+5)-x*2-5x=0
VT=-5(x-2)
<=>-5(x-2)=0
<=>x-2=0
<=>x=2
c)(2x-3)*2-(x-5)*2=0
VT=2(x+2)
<=>2(x+2)=0
<=>x+2=0
<=>x=-2
d) 3x*3-48x=0
VT=-39x
<=>-39x=0
<=>x=0
e) x*3+x*2-4x=4
VT=x
<=>x=4
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=-x^2+3x-5\)\(=-\dfrac{11}{4}-\left(x^2-2.\dfrac{3}{2}x+\dfrac{9}{4}\right)=-\dfrac{11}{4}-\left(x-\dfrac{3}{2}\right)^2\le-\dfrac{11}{4}\) với mọi x
\(\Rightarrow A_{max}=-\dfrac{11}{4}\Leftrightarrow x-\dfrac{3}{2}=0\Leftrightarrow x=\dfrac{3}{2}\)
\(B=5x-4x^2-3=-\dfrac{23}{16}-\left(4x^2-2.\dfrac{5}{4}.2x+\dfrac{25}{16}\right)\)\(=-\dfrac{23}{16}-\left(2x-\dfrac{5}{4}\right)^2\)\(\le-\dfrac{23}{16}\forall x\)
\(\Rightarrow B_{max}=-\dfrac{23}{16}\Leftrightarrow2x-\dfrac{5}{4}=0\Leftrightarrow x=\dfrac{5}{8}\)
\(C=5-4x-25x^2=\dfrac{129}{25}-\left(25x^2+2.5x.\dfrac{2}{5}+\dfrac{4}{25}\right)\)\(=\dfrac{129}{25}-\left(5x+\dfrac{2}{5}\right)^2\le\dfrac{129}{25}\forall x\)
\(\Rightarrow C_{max}=\dfrac{129}{25}\Leftrightarrow5x+\dfrac{2}{5}=0\Leftrightarrow x=-\dfrac{2}{25}\)
\(D=3x-2x^2=-2\left(x^2-\dfrac{3}{2}x\right)=-2\left(x^2-2.\dfrac{3}{4}x+\dfrac{9}{16}\right)+\dfrac{9}{8}\)\(=\dfrac{9}{8}-2\left(x-\dfrac{3}{4}\right)^2\le\dfrac{9}{8}\) với mọi x
\(\Rightarrow D_{max}=\dfrac{9}{8}\Leftrightarrow x-\dfrac{3}{4}=0\Leftrightarrow x=\dfrac{3}{4}\)
\(E=2+6x-\dfrac{1}{4}x^2=-\dfrac{1}{4}\left(x^2-24x\right)+2=-\dfrac{1}{4}\left(x^2-2.12x+144\right)+38\)\(=38-\dfrac{1}{4}\left(x-12\right)^2\le38\forall x\)
\(\Rightarrow E_{max}=38\Leftrightarrow x-12=0\Leftrightarrow x=12\)
\(F=-5x^2+4x=-5\left(x^2-\dfrac{4}{5}x\right)=-5\left(x^2-2.\dfrac{2}{5}x+\dfrac{4}{25}\right)+\dfrac{4}{5}\)\(=\dfrac{4}{5}-5\left(x-\dfrac{2}{5}\right)^2\le\dfrac{4}{5}\forall x\)
\(\Rightarrow F_{max}=\dfrac{4}{5}\Leftrightarrow x-\dfrac{2}{5}=0\Leftrightarrow x=\dfrac{2}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left|x^2-3x+1\right|=x+1\)
Ta có:
TH1: \(x^2-3x+1=x+1\Rightarrow x^2-3x+1-\left(x+1\right)=0\)
\(\Rightarrow x^2-3x+1-x-1=0\Rightarrow x^2-4x=0\)
\(\Rightarrow x\left(x-4\right)=0\Rightarrow\hept{\begin{cases}x=0\\x-4=0\end{cases}}\Rightarrow\hept{\begin{cases}x=0\\x=4\end{cases}}\)
TH2: \(x^2-3x+1=-\left(x+1\right)=-x-1\left(1\right)\)
\(\Rightarrow x^2-3x+1-\left(-x-1\right)=0\Rightarrow x^2-3x+1+x+1=0\)
\(\Rightarrow x^2-2x+2=0\Rightarrow\left(x^2-2x+1\right)+1=0\Rightarrow\left(x^2-2.x.1+1^2\right)+1=0\)
\(\Rightarrow\left(x-1\right)^2+1=0\)
Vì \(\left(x-1\right)^2\ge0\Rightarrow\left(x-1\right)^2+1\ge1>0\)
=>PT (1) vô nghiệm
Vậy \(x=0;x=4\) là nghiệm của PT
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a)=\(3x^3-15x^2+21x\)
b)\(=-2x^4y-10x^2y+2xy\)
c)\(=-x^3+6x^2+5x-4x^2+24x+20=-x^3+2x^2+29x+20\)
d)\(=2x^4-3x^3+4x^2-2x^2+3x-4=2x^4-3x^32x^2+3x-4\)
e)\(=x^2-4y^2\)
f)\(=-2x^2y^3+y-3\)
g)\(=3xy^4-\dfrac{1}{2}y^2+2x^2y\)
h)\(=9x^2-6x+1-7x^2-14=2x^2-6x-13\)
i)\(=x^2-x-3\)
j)\(=\left(x+2y\right)\left(x^2-2y+4y^2\right):\left(x+2y\right)=x^2-2y+4y^2\)
\(5x^2\)\(-15x+5+x-5x^2\)\(=x-2\)
\(-15x+5=-2\)
\(-15x=-2-5=-7\)
\(x=-7:-15=\frac{-7}{-15}\)\(=\frac{7}{15}\)
\(x=\frac{7}{15}\)