A=\(|x-9|+7\) khi \(x\ge9\)
B=\(|-3x|-8x^2+x-2\) khi \(x< 0\)
C=\(3x+|x-1|-2\)
D=\(\dfrac{x^2-|2x|+1}{\left(x+1\right)|-x|}\) khi x<0
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a: =>x^2+4x-4x+1=0
=>x^2+1=0
=>Loại
b: =>2x-6+4=2x+2
=>-2=2(loại)
c: =>2(x+3)-2x-1=1
=>6-1=1
=>5=1(loại)
d =>x+3=0
=>x=-3(loại)
e: =>x^2-3x^2+3x-3x-2=0
=>-2x^2-2=0
=>x^2+1=0
=>Loại
câu d
\(D=\dfrac{\left(1-x^2\right)}{x}\left(\dfrac{x^2}{x+3}-1\right)+\dfrac{3x^2-14x+3}{x^2+3x}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{\left(1-x^2\right)\left(x^2-x-3\right)+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{x^2-x-3-x^4+x^3-3x^2+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x^4+x^3+x^2-15x}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x\left(x^3-x^2-x+15\right)}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-\left(x^3-x^2-x+15\right)}{\left(x+3\right)}\end{matrix}\right.\)
Với \(x=0\) không phải nghiệm
Với \(x\ne0\) chia 2 vế cho \(x^2\), pt tương đương:
\(2x^2+3x-1+\dfrac{3}{x}+\dfrac{2}{x^2}=0\)
\(\Leftrightarrow2\left(x+\dfrac{1}{x}\right)^2+3\left(x+\dfrac{1}{x}\right)-5=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=1\\x+\dfrac{1}{x}=-\dfrac{5}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+1=0\\2x^2+5x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vô-nghiệm\right)\\\left(x+2\right)\left(2x+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Câu a chắc là đề sai, vì nghiệm vô cùng xấu, tử số của phân thức cuối cùng là \(x+17\) mới hợp lý
b.
Đặt \(x+3=t\)
\(\Rightarrow\left(t+1\right)^4+\left(t-1\right)^4=14\)
\(\Leftrightarrow t^4+6t^2-6=0\) (đến đây đoán rằng bạn tiếp tục ghi sai đề, nhưng thôi cứ giải tiếp)
\(\Rightarrow\left[{}\begin{matrix}t^2=-3+\sqrt{15}\\t^2=-3-\sqrt{15}\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow t=\pm\sqrt{-3+\sqrt{15}}\Rightarrow x=-3\pm\sqrt{-3+\sqrt{15}}\)
Câu c chắc cũng sai đề, vì lên lớp 8 rồi không ai cho đề kiểu này cả, người ta sẽ rút gọn luôn số 1 bên trái và 60 bên phải.
a: =>10x-4=15-9x
=>19x=19
hay x=1
b: \(\Leftrightarrow3\left(10x+3\right)=36+4\left(8x+6\right)\)
=>30x+9=36+32x+24
=>30x-32x=60-9
=>-2x=51
hay x=-51/2
c: \(\Leftrightarrow2x+\dfrac{6}{5}=5-\dfrac{13}{5}-x\)
=>3x=6/5
hay x=2/5
d: \(\Leftrightarrow\dfrac{7x}{8}-\dfrac{5\left(x-9\right)}{1}=\dfrac{20x+1.5}{6}\)
\(\Leftrightarrow21x-120\left(x-9\right)=4\left(20x+1.5\right)\)
=>21x-120x+1080=80x+60
=>-179x=-1020
hay x=1020/179
e: \(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)
=>35x-5+60x=96-6x
=>95x+6x=96+5
=>x=1
f: \(\Leftrightarrow6\left(x+4\right)+30\left(-x+4\right)=10x-15\left(x-2\right)\)
=>6x+24-30x+120=10x-15x+30
=>-24x+96=-5x+30
=>-19x=-66
hay x=66/19
b: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
hay \(x\in\left\{-\dfrac{10}{7};3\right\}\)
d: \(\Leftrightarrow\dfrac{13}{2x^2+7x-6x-21}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow26x+91+x^2-9-12x-14=0\)
\(\Leftrightarrow x^2+14x+68=0\)
hay \(x\in\varnothing\)
`a,` \(\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)
`<=> (5(5x+2))/30 - (10(8x-1))/30 = (6(4x+2))/30 - (5.30)/30`
`<=> 5(5x+2) - 10(8x-1) =6(4x+2) - 5.30`
`<=> 25x + 10 - 80x + 10 = 24x+12 - 150`
`<=> -55x +20 = 24x-138`
`<=> -55x -24x=-138-20`
`<=>-79x=-158`
`<=> x=2`
Vậy pt có nghiệm `x=2`
`b,` \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\)
ĐKXĐ : \(\left\{{}\begin{matrix}x-2\ne0\\x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne2\\x\ne0\end{matrix}\right.\)
Ta có : `(x+2)/(x-2) -1/x = 2/(x(x-2))`
`<=> (x(x+2))/(x(x-2)) - (x-2)/(x(x-2)) = 2/(x(x-2))`
`=> x^2 +2x - x +2 = 2`
`<=> x^2 + x =0`
`<=>x(x+1)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=-1\end{matrix}\right.\)
Vậy pt có nghiệm `x=-1`
`c,2x^3 + 6x^2 =x^2 +3x`
`<=> 2x^3 + 6x^2 -x^2 -3x=0`
`<=> 2x^3 + 5x^2 -3x=0`
`->` Đề có sai ko ạ ?
`d,` \(\left|x-4\right|+3x=5\) `(1)`
Thường hợp `1` : `x-4 >= 0<=> x >=0` thì phương trình `(1)` thở thành :
`x-4 = 5-3x`
`<=> x+3x=5+4`
`<=> 4x=9`
`<=> x= 9/4 (t//m)`
Trường hợp `2` : `x-4< 0<=> x<0` thì phương trình `(1)` trở thành :
`-(x-4) =5-3x`
`<=> -x +4=5-3x`
`<=> -x+3x=5-4`
`<=> 2x =1`
`<=>x=1/2 ( kt//m)`
Vậy phương trình có nghiệm `x=9/4`
a: =>2x-1=-2
=>2x=-1
hay x=-1/2
b: \(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\-\dfrac{2}{5}x-7=0\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{2}{3};-\dfrac{35}{2}\right\}\)
c: x/8=9/4
nên x/8=18/8
hay x=18
d: \(\Leftrightarrow\left(x-3\right)^2=36\)
=>x-3=6 hoặc x-3=-6
=>x=9 hoặc x=-3
e: =>-1,7x=6,12
hay x=-3,6
h: =>x-3,4=27,6
hay x=31
a) \(\dfrac{1}{3}\div\left(2x-1\right)=\dfrac{-1}{6}\)
\(\left(2x-1\right).\dfrac{1}{3}\div\left(2x-1\right)=\left(2x-1\right)\left(-\dfrac{1}{6}\right)\)
\(\dfrac{1}{3}=\left(2x-1\right)\left(-\dfrac{1}{6}\right)\)
\(\dfrac{1}{3}=-1\left(2x-1\right)\div6\)
\(\dfrac{1}{3}=-2x+1\div6\)
\(x=-\dfrac{1}{2}\)
b) \(\left(3x+2\right)\left(\dfrac{-2}{5}x-7\right)=0\)
\(TH1:3x+2=0\)
\(3x=0-2\)
\(3x=-2\)
\(x=\dfrac{-2}{3}\)
\(TH2:\left(-\dfrac{2}{5}x-7\right)=0\)
\(\left(\dfrac{-2}{5}x-7\right)=0\)
\(\left(\dfrac{-2x}{5}+\dfrac{5\left(-7\right)}{5}\right)=0\)
\(\left(\dfrac{-2x-35}{5}\right)=0\)
\(-2x-35=0\)
\(-2x=0+35\)
\(x=-\dfrac{35}{2}\)
c) \(\dfrac{x}{8}=\dfrac{9}{4}\)
\(\Leftrightarrow x=\dfrac{9.8}{4}=\dfrac{72}{4}=18\)
\(x=18\)
d) \(\dfrac{x-3}{2}=\dfrac{18}{x-3}\)
\(x-3=18+2\)
\(x=20-3\)
\(x=17\)
e) \(4,5x-6,2x=6,12\)
\(\dfrac{9x}{2}-6,2.x=6,12\)
\(\dfrac{9x}{2}+\dfrac{-31x}{5}=6,12\)
\(\dfrac{5.9x}{10}+\dfrac{2\left(-31\right)x}{10}=6.12\)
\(\dfrac{45x-62x}{10}=6.12\)
\(=-17x\div10=6.12\)
\(-17x=10.6.12\)
\(x=-3,6\)
h) \(11,4-\left(x-3,4\right)=-16,2\)
\(x-3,4=-16,2+11,4\)
\(x-3,4=-4,8\)
\(x=-1,4\)
a) Ta có: \(\left(2x-3\right)^2=\left(2x-3\right)\left(x+1\right)\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2x-3-x-1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=4\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{3}{2};4\right\}\)
b) Ta có: \(x\left(2x-9\right)=3x\left(x-5\right)\)
\(\Leftrightarrow x\left(2x-9\right)-3x\left(x-5\right)=0\)
\(\Leftrightarrow x\left(2x-9\right)-x\left(3x-15\right)=0\)
\(\Leftrightarrow x\left(2x-9-3x+15\right)=0\)
\(\Leftrightarrow x\left(6-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
Vậy: S={0;6}
c) Ta có: \(3x-15=2x\left(x-5\right)\)
\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{5;\dfrac{3}{2}\right\}\)
d) Ta có: \(\dfrac{5-x}{2}=\dfrac{3x-4}{6}\)
\(\Leftrightarrow6\left(5-x\right)=2\left(3x-4\right)\)
\(\Leftrightarrow30-6x=6x-8\)
\(\Leftrightarrow30-6x-6x+8=0\)
\(\Leftrightarrow-12x+38=0\)
\(\Leftrightarrow-12x=-38\)
\(\Leftrightarrow x=\dfrac{19}{6}\)
Vậy: \(S=\left\{\dfrac{19}{6}\right\}\)
e) Ta có: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)
\(\Leftrightarrow6x+4-3x-1=12x+10\)
\(\Leftrightarrow3x+3-12x-10=0\)
\(\Leftrightarrow-9x-7=0\)
\(\Leftrightarrow-9x=7\)
\(\Leftrightarrow x=-\dfrac{7}{9}\)
Vậy: \(S=\left\{-\dfrac{7}{9}\right\}\)
a: =>1/3x+2/5x-2/5=0
=>11/15x-2/5=0
=>11/15x=2/5
=>x=2/5:11/15=2/5*15/11=30/55=6/11
b: =>-5x-1-1/2x+1/3=x
=>-11/2x-2/3-x=0
=>-13/2x=2/3
=>x=-2/3:13/2=-2/3*2/13=-4/39
c: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=1/3 hoặc x=-1/2
d: 9(3x+1)^2=16
=>(3x+1)^2=16/9
=>3x+1=4/3 hoặc 3x+1=-4/3
=>3x=1/3 hoặc 3x=-7/3
=>x=1/9 hoặc x=-7/9