Tìm x
\(a,3x-\left|2x+1\right|=2\)
\(b,\left(x+7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
A)\(^{ }\left(^{ }x+1\right)\left(2x-1\right)\left(-x+2\right)=0\)
B)\(^{ }\left(2x-1\right)\left(3x+2\right)\left(4x-5\right)\left(x-7\right)=0\)
C)\(^{ }x^2-6x+11=0\)
D)(\(\left(x^2+2x+3\right)\left(x^2-25\right)\left(x+19\right)=0\)
a) \(\left(x+1\right)\left(2x-1\right)\left(-x+2\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x+1=0\\2x-1=0\\-x+2=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-1\\x=\frac{1}{2}\\x=2\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{-1;\frac{1}{2};2\right\}\)
b) \(\left(2x-1\right)\left(3x+2\right)\left(4x-5\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}2x-1=0\\3x+2=0\\4x-5=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=\frac{1}{2}\\x=-\frac{2}{3}\\x=\frac{5}{4}\\x=7\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{\frac{1}{2};-\frac{2}{3};\frac{5}{4};7\right\}\)
c) \(x^2-6x+11=0\)
\(\Leftrightarrow x^2-6x+9+2=0\)
\(\Leftrightarrow\left(x-3\right)^2+2=0\) (vô lí)
Vậy phương trình vô nghiệm
d) \(\left(x^2+2x+3\right)\left(x^2-25\right)\left(x+19\right)=0\)
\(\Leftrightarrow\left(x^2+2x+1+2\right)\left(x+5\right)\left(x-5\right)\left(x+19\right)=0\)
\(\Leftrightarrow\left[\left(x+1\right)^2+2\right]\left(x+5\right)\left(x-5\right)\left(x+19\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x+5=0\\x-5=0\\x+19=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-5\\x=5\\x=-19\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{\pm5;-19\right\}\)
a,b,d dễ mà bạn tự làm
c,x2-6x+11=0<=> x2-6x+9+2=0
<=>(x-3)2=-2(vô lý)
vậy pt vô nghiệm
chứng minh rằng các biểu thức sau không phụ thuộc vào x:
a. \(A=\left(3x+7\right)\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\)
b. \(B=\left(x^2-2\right)\left(x^2+x-1\right)-x\left(x^3+x^2-3x-2\right)\)
c. \(C=x\left(x^3+x^2-3x-2\right)-\left(x^2-2\right)\left(x^2+x-1\right)\)
A)\(\left(2x-7\right)^2-6\left(2x-7\right)\left(x-3\right)=0\)
B) \(x^3+1+\left(x^2-x+1\right)=0\)
C) \(\left(3x+1\right)^2-x^2+8x-16=0\)
D)\(\left(x+1\right)\left(x-1\right)^2\left(x+1\right)\left(x-2\right)^2=0\)
E) \(\left(3x+1\right)\left(x-3\right)^2=\left(3x+1\right)\left(2x-5\right)^2\)
F) \(\left(x+5\right)\left(3x+2\right)^2=x^2\left(x+5\right)\)
BÀI 6 tìm x
1,\(2x\left(x-5\right)-\left(3x+2x^2\right)=0\) 2,\(x\left(5-2x\right)+2x\left(x-1\right)=13\)
3,\(2x^3\left(2x-3\right)-x^2\left(4x^2-6x+2\right)=0\) 4,\(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
5,\(6x^2-\left(2x-3\right)\left(3x+2\right)=1\) 6,\(2x\left(1-x\right)+5=9-2x^2\)
1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)
=>-13x=0
=>x=0
2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
=>3x=13
=>x=13/3
3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)
=>-2x^2=0
=>x=0
4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
=>-8x=6-14=-8
=>x=1
`1)2x(x-5)-(3x+2x^2)=0`
`<=>2x^2-10x-3x-2x^2=0`
`<=>-13x=0`
`<=>x=0`
___________________________________________________
`2)x(5-2x)+2x(x-1)=13`
`<=>5x-2x^2+2x^2-2x=13`
`<=>3x=13<=>x=13/3`
___________________________________________________
`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`
`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`
`<=>x=0`
___________________________________________________
`4)5x(x-1)-(x+2)(5x-7)=0`
`<=>5x^2-5x-5x^2+7x-10x+14=0`
`<=>-8x=-14`
`<=>x=7/4`
___________________________________________________
`5)6x^2-(2x-3)(3x+2)=1`
`<=>6x^2-6x^2-4x+9x+6=1`
`<=>5x=-5<=>x=-1`
___________________________________________________
`6)2x(1-x)+5=9-2x^2`
`<=>2x-2x^2+5=9-2x^2`
`<=>2x=4<=>x=2`
Tìm x biết
1) \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
2)\(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x+1\right)-33\)
3)\(6x\left(3x+5\right)-2x\left(9x-2\right)+\left(17-x\right)\left(x-1\right)+x\left(x-18\right)-17x^2=0\)
4)\(\left(x-1\right)\left(x+2\right)-\left(x-3\right)+5x-7=0\)
Giúp mình nha. Camon nhiều
Tìm x
a, \(9x^2-49=0\) b, \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-1\right)-27=0\)
c, (x-1)(x+2)-x-2=0 d, \(x\left(3x+2\right)+\left(x+1\right)^2-\left(2x-5\right)\left(2x+5\right)=0\)
e, (4x+1)(x-2)-(2x-3)(2x+1)=7
Giair phương trình sau:
a,\(2x^3+5x^2-3x=0\) b,\(2x^3+6x^2=x^2+3x\)
c,\(x^2+\left(x+2\right)\left(11x-7\right)=4\) d,\(\left(x-1\right)\left(x^2+5x-2\right)-\left(x^3-1\right)=0\)
e, \(x^3+1=x\left(x+1\right)\) f,\(x^3+x^2+x+1=0\)
g,\(x^3-3x^2+3x-1=0\) h,\(x^3-7x+6=0\)
i,\(x^6-x^2=0\) j,\(x^3-12=13x\)
k,\(-x^5+4x^4=-12x^3\) l, \(x^3=4x\)
a) Ta có: \(2x^3+5x^2-3x=0\)
\(\Leftrightarrow x\left(2x^2+5x-3\right)=0\)
\(\Leftrightarrow x\left(2x^2+6x-x-3\right)=0\)
\(\Leftrightarrow x\left[2x\left(x+3\right)-\left(x+3\right)\right]=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)
b) Ta có: \(2x^3+6x^2=x^2+3x\)
\(\Leftrightarrow2x^2\left(x+3\right)=x\left(x+3\right)\)
\(\Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)
c) Ta có: \(x^2+\left(x+2\right)\left(11x-7\right)=4\)
\(\Leftrightarrow x^2+11x^2-7x+22x-14-4=0\)
\(\Leftrightarrow12x^2+15x-18=0\)
\(\Leftrightarrow12x^2+24x-9x-18=0\)
\(\Leftrightarrow12x\left(x+2\right)-9\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(12x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\12x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\12x=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{-2;\dfrac{3}{4}\right\}\)
Trong đó có nhiều phương trình kiến thức cơ bản mà nhỉ? Ít nâng cao, bạn lọc ra câu nào k làm đc thôi chứ!
Tính giá trị biểu thức?
\(A=\left(3x+7\right)\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\)\(\)
\(B=\left(x+1\right)\left(x^2-x-1\right)-\left(x-1\right)\left(x^2+x+1\right)\)
A=(3x+7)(2x+3)-(3x-5)(2x+11) =6x2+9x+14x+21-6x2-33x+10x+55 =(6x2-6x2)+(9x+14x-33x+10x)+(21+55) =76
\(A=\left(3x+7\right)\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\)
\(\Leftrightarrow A=6x^2+14x+9x+21-\left(6x^2-10x+33x-55\right)\)
\(\Leftrightarrow A=6x^2+23x+21-\left(6x^2+23x-55\right)\)
\(\Leftrightarrow A=6x^2+23x+21-6x^2-23x+55\)
\(\Leftrightarrow A=76\)
\(B=\left(x+1\right)\left(x^2-x-1\right)-\left(x-1\right)\left(x^2+x+1\right)\)
\(\Leftrightarrow B=\left(x+1\right)x^2-x\left(x+1\right)-\left(x+1\right)-\left(x-1\right)x^2-\left(x-1\right)x-\left(x-1\right)\)
\(\Leftrightarrow B=x^3+x^2-x^2-x-x-1-x^3+x^2-x^2+x-x+1\)
\(\Leftrightarrow B=\left(x^3-x^3\right)+\left(x^2-x^2+x^2-x^2\right)+\left(x-x-x-x\right)+\left(1-1\right)\)
\(\Leftrightarrow B=-2x\)
Bài 11 : Tìm GTNN của của các biểu thức sau :
a ) \(A=\left|x+3\right|+\left|2x-5\right|+\left|x-7\right|.\)
b ) \(B=\left|x+2\right|+\left|3x-4\right|+\left|x-2\right|+5\)
c ) \(M=\left|x+2\right|+\left|x-3\right|\)
d ) \(C=\left|2x+5\right|+\left|2x+1\right|+\left|2x-7\right|+\left|2x-4\right|+4\)
e ) \(D=\left|3x-6\right|+\left|3x-9\right|+\left|3x-12\right|+\left|3x-15\right|+2018\)