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\(8,1-\left(x-6\right)=4\left(2-2x\right)\)
\(\Leftrightarrow1-x+6=8-8x\)
\(\Leftrightarrow-x+8x=8-1-6\)
\(\Leftrightarrow7x=1\)
\(\Leftrightarrow x=\dfrac{1}{7}\)
\(9,\left(3x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)
\(10,\left(x+3\right)\left(x^2+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\varnothing\end{matrix}\right.\)
`8)1-(x-5)=4(2-2x)`
`<=>1-x+5=8-6x`
`<=>5x=2<=>x=2/5`
`9)(3x-2)(x+5)=0`
`<=>[(x=2/3),(x=-5):}`
`10)(x+3)(x^2+2)=0`
Mà `x^2+2 > 0 AA x`
`=>x+3=0`
`<=>x=-3`
`11)(5x-1)(x^2-9)=0`
`<=>(5x-1)(x-3)(x+3)=0`
`<=>[(x=1/5),(x=3),(x=-3):}`
`12)x(x-3)+3(x-3)=0`
`<=>(x-3)(x+3)=0`
`<=>[(x=3),(x=-3):}`
`13)x(x-5)-4x+20=0`
`<=>x(x-5)-4(x-5)=0`
`<=>(x-5)(x-4)=0`
`<=>[(x=5),(x=4):}`
`14)x^2+4x-5=0`
`<=>x^2+5x-x-5=0`
`<=>(x+5)(x-1)=0`
`<=>[(x=-5),(x=1):}`
a: =>|x-7|=3-2x
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(-2x+3\right)^2-\left(x-7\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(2x-3-x+7\right)\left(2x-3+x-7\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(x+4\right)\left(3x-10\right)=0\end{matrix}\right.\Leftrightarrow x=-4\)
b: =>|2x-3|=4x+9
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(4x+9-2x+3\right)\left(4x+9+2x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(2x+12\right)\left(6x+6\right)=0\end{matrix}\right.\Leftrightarrow x=-1\)
c: =>3x+5=2-5x hoặc 3x+5=5x-2
=>8x=-3 hoặc -2x=-7
=>x=-3/8 hoặc x=7/2
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
4(18-5x)-12(3x-7)=15(2x-16)-6(x+14)
<=>72-20x-36x+84=30x-240x-6x-84
<=>160x=-86
<=>x=-0.0375
Bài 3:
a) \(\left(x-6\right).\left(2x-5\right).\left(3x+9\right)=0\)
\(\Leftrightarrow\left(x-6\right).\left(2x-5\right).3.\left(x+3\right)=0\)
Vì \(3\ne0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\2x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\2x=5\\x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\frac{5}{2}\\x=-3\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{6;\frac{5}{2};-3\right\}.\)
b) \(2x.\left(x-3\right)+5.\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right).\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\frac{5}{2}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{3;-\frac{5}{2}\right\}.\)
c) \(\left(x^2-4\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x^2-2^2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2-3+2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2;\frac{1}{3}\right\}.\)
Chúc bạn học tốt!
a) Chỗ sai trong phương trình là: \(5 - x + 8 = 3x + 3x - 27\) (dòng thứ 2) vì khi phá ngoặc đã không đổi dấu của số 8.
Sửa lại:
\(\begin{array}{l}5 - \left( {x + 8} \right) = 3x + 3\left( {x - 9} \right)\\\,\,\,\,5 - x - 8 = 3x + 3x - 27\\\,\,\,\,\,\,\, - 3 - x = 6x - 27\\\,\,\,\, - x - 6x = - 27 + 3\\\,\,\,\,\,\,\,\,\,\,\,\, - 7x = - 24\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = \left( { - 24} \right):\left( { - 7} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = \frac{{24}}{7}\end{array}\)
Vậy phương trình có nghiệm \(x = \frac{{24}}{7}.\)
b) Chỗ sai trong phương trình là: \(4x + 5x = 9 - 18\) (dòng thứ 3) vì khi chuyển \( - 18\) từ vế trái sang vế phải đã không đổi dấu thành \( + 18\).
Sửa lại:
\(\begin{array}{l}3x - 18 + x = 12 - \left( {5x + 3} \right)\\\,\,\,\,\,\,\,4x - 18 = 12 - 5x - 3\\\,\,\,\,\,\,\,4x + 5x = 9 + 18\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,9x = 27\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 27:9\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 3.\end{array}\)
Vậy phương trình có nghiệm \(x = 3.\)
a: \(\Leftrightarrow x^3-3x^2+3x-1-x^3+2x^2-x=5x\left(2-x\right)-11\left(x+2\right)\)
=>-x^2+2x-1=10x-5x^2-11x-22
=>-x^2+2x-1=-5x^2-x-22
=>4x^2+3x+21=0
=>PTVN
b: \(\Leftrightarrow\left(x+10\right)\left(x+4\right)+3\left(x+4\right)\left(x-2\right)=4\left(x+10\right)\left(x-2\right)\)
=>x^2+14x+40+3(x^2+2x-8)=4(x^2+8x-20)
=>x^2+14x+40+3x^2+6x-24=4x^2+32x-80
=>20x+16=32x-80
=>-12x=-96
=>x=8
c: \(\Leftrightarrow6\left(x-3\right)+7\left(x-5\right)=13x+4\)
=>6x-18+7x-35=13x+4
=>-53=4(loại)
d: =>3(2x-1)-5(x-2)=3(x+7)
=>6x-3-5x+10=3x+21
=>3x+21=x+7
=>x=-7
e: =>x^3-6x^2+12x-8-x^3-3x^2-3x-1=-9x^2+1
=>-9x^2+9x-9=-9x^2+1
=>9x=10
=>x=10/9
Bài1:
\(a,\left(-8\right)^9\) và \(\left(-32\right)^5\)
Ta có:
\(\left(-8\right)^9=-2^{27}\)
\(\left(-32\right)^5=\left(-8.4\right)^5=-2^{27}.2^{10}\)
Vì \(-2^{27}.10< -2^{27}\) nên \(\left(-8\right)^9>\left(-32\right)^5\)
Các câu sau tương tự
Bài2:
\(a,2\left|x-1\right|-3x=7\)
+)Xét \(x\ge1\Rightarrow\left|x-1\right|=x-1\)
Do đó:
\(2\left(x-1\right)-3x=7\\ \Leftrightarrow2x-2-3x=7\\ \Leftrightarrow-x=9\\ \Leftrightarrow x=-9\left(loại\right)\)
+)Xét \(x< 1\Rightarrow\left|x-1\right|=1-x\)
Do đó:
\(2\left(1-x\right)-3x=7\\ \Leftrightarrow2-2x-3x=7\\ \Leftrightarrow-5x=5\\ x=-1\left(chon\right)\)
Vậy x=-1
Câu b tương tự
Bài 1:
\(a,\left(-8\right)^9\) và \(\left(-32\right)^5\)
\(\left(-8\right)^9=\left[\left(-2\right)^3\right]^9=\left(-2\right)^{27}\)
\(\left(-32\right)^5=\left[\left(-2\right)^5\right]^5=\left(-2\right)^{25}\)
\(\left(-2\right)^{27}< \left(-2\right)^{25}\)
\(\Rightarrow\left(-8\right)^9< \left(-32\right)^5\)
\(b,2^{21}\) và \(3^{14}\)
\(2^{21}=\left(2^3\right)^7\)
\(3^{14}=\left(3^2\right)^7\)
\(2^3< 3^2\)\(\Rightarrow2^{21}< 3^{14}\)
\(c,12^8\) và \(8^{12}\)
\(12^8=\left(12^2\right)^4=144^4\)
\(8^{12}=\left(8^3\right)^4=512^4\)
\(144^4< 512^4\)\(\Rightarrow12^8< 8^{12}\)
\(d,\left(-5\right)^{39}\) và \(\left(-2\right)^{91}\)
\(\left(-5\right)^{39}=\left[\left(-5\right)^3\right]^{13}\)
\(\left(-2\right)^{91}=\left[\left(-2\right)^7\right]^{13}\)
\(\left(-5\right)^3>\left(-2\right)^7\)\(\Rightarrow\left(-5\right)^{39}>\left(-2\right)^{91}\)
Bài 2:
\(a,2.\left|x-1\right|-3x=7\)
\(\left|x-1\right|=\dfrac{7+3x}{2}\)
Ta có 2 trường hợp:
Th1:\(x-1=\dfrac{7-3x}{2}\)
\(\dfrac{2x-2}{2}=\dfrac{7+3x}{2}\)
\(\Rightarrow2x-2=7+3x\)
\(2x-3x=7+2\)
\(-x=9\Rightarrow x=-9\)
Th2:\(x+1=-\dfrac{7+3x}{2}\)
\(\dfrac{2x-2}{2}=\dfrac{-7-3x}{2}\)
\(\Rightarrow2x-2=-7-3x\)
\(2x+3x=-7+2\)
\(5x=-5\Rightarrow x=-1\)
Vậy \(x\in\left\{-9;-1\right\}\)
\(b,\left|5x-3\right|=\left|7-x\right|\)
Ta có: Th1: \(\left|7-x\right|=7-x\) khi \(7-x\ge0\)\(\Rightarrow x\le7\)
\(5x-3=7-x\)
\(5x+x=7+3\)
\(6x=10\Rightarrow x=\dfrac{10}{6}=\dfrac{5}{3}\)( thoả mãn )
vì x thoả mãn \(x\le7\)\(\Rightarrow\) th1 thoả mãn x
Ta có: Th2: \(\left|7-x\right|=-\left(7-x\right)\) khi \(7-x< 0\Rightarrow x>7\)
\(5x-3=-\left(7-x\right)\)
\(5x-3=-7+x\)
\(5x-x=-7+3\)
\(4x=-4\Rightarrow x=-1\) ( loại )
Vì x thoả mãn \(x>7\) mà \(x=-1\Rightarrow\)th2 loại