a.  -12 (x-5) +7(3-x) = -12x+60+21-7x=-19x+81=15

=> -19x=15-81=-66

=> x=66/19

b. 30(x+2) - 5 (x-5) - 24x = 30x+60-5x+25-24x =(30x-5x-24x)+(60+25)=x+85=100

=> x=100-85=15

c. ( 30x + 60 ) - ( 6x - 30 ) - 24x = 100

=30x+60-6x+30-24x

=90=100 (cô lý)

Vậy không có x thỏa mãn

1500: [(30x + 40) : x]=30

=>(30x + 40): x = 1500 : 30

=>30x : x+ 40:x=50

=> 30 + 40 :x=50

=> 40 :x=50-30

=> 40:x=20

=>x=40:20

=>x=2

 : [(30x + 40) : x] = 30(2x - 15)5 = (2x - 15)3

=>(30x+40):x=1500 : 30=50

40:x = 50-30

40:x=20

x=2

b)\(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)-24\)4

\(=\left[\left(x-1\right)\left(x-4\right)\right]\left[\left(x-2\right)\left(x-3\right)\right]-24\)

\(=\left(x^2-4x-x+4\right)\left(x^2-3x-2x+6\right)-24\)

\(=\left(x^2-5x+4\right)\left(x^2-5x+4+2\right)-24\)

\(\)Đặt  \(x^2-5x+4\)là a,ta có

\(=a\left(a+2\right)-24\)

\(=a^2+2a-24\)

\(=a^2+6a-4a-24\)

\(=a\left(a+6\right)-4\left(a+6\right)\)

\(=\left(a+6\right)\left(a-4\right)\)

Hay  \(\left(x^2-5x+4+6\right)\left(x^2-5x+4-4\right)\)

\(=\left(x^2-5x+10\right)\left(x^2-5\right)\)

Câu hỏi của Huỳnh Bảo Nguyên - Toán lớp 8 - Học toán với OnlineMath

Mk làm òi nhé !

a) Ta có: \(\frac{x-1}{2012}+\frac{x-2}{2011}+\frac{x-3}{2010}+...+\frac{x-2012}{1}=2012\)

\(\Leftrightarrow\frac{x-1}{2012}+\frac{x-2}{2011}+\frac{x-3}{2010}+...+\frac{x-2012}{1}-2012=0\)

\(\Leftrightarrow\frac{x-1}{2012}-1+\frac{x-2}{2011}-1+\frac{x-3}{2010}-1+...+\frac{x-2012}{1}-1=0\)

\(\Leftrightarrow\frac{x-2013}{2012}+\frac{x-2013}{2011}+\frac{x-2013}{2010}+...+\frac{x-2013}{1}=0\)

\(\Leftrightarrow\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+1\right)=0\)

mà \(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+1>0\)

nên x-2013=0

hay x=2013

Vậy: Tập nghiệm S={2013}

b) Ta có: \(x^4-30x^2+31x-30=0\)

\(\Leftrightarrow x^4+x-30x^2+30x-30=0\)

\(\Leftrightarrow\left(x^4+x\right)-\left(30x^2-30x+30\right)=0\)

\(\Leftrightarrow x\left(x^3+1\right)-30\left(x^2-x+1\right)=0\)

\(\Leftrightarrow x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left[x\left(x+1\right)-30\right]=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(x^2+x-30\right)=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(x^2+6x-5x-30\right)=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left[x\left(x+6\right)-5\left(x+6\right)\right]=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(x+6\right)\left(x-5\right)=0\)(1)

Ta có: \(x^2-x+1\)

\(=x^2-2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)
\(=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\)

Ta có: \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)

hay \(x^2-x+1>0\forall x\)(2)

Từ (1) và (2) suy ra (x+6)(x-5)=0

\(\Leftrightarrow\left[{}\begin{matrix}x+6=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=5\end{matrix}\right.\)

Vậy: Tập nghiệm S={-6;5}

a)

PT <=> \(\left(\frac{x-1}{2012}-1\right)+\left(\frac{x-2}{2011}-1\right)+...+\left(\frac{x-2012}{1}-1\right)=0\)

<=> \(\frac{x-2013}{2012}+\frac{x-2013}{2011}+...+\frac{x-2013}{1}=0\)

<=> \(\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+...+\frac{1}{1}\right)=0\)

Mà \(\frac{1}{2012}+\frac{1}{2011}+...+\frac{1}{1}\ne0\)

<=> x - 2013 = 0

<=> x = 2013

KL: ...

b) PT <=> \(\left(x^4-5x^3\right)+\left(5x^3-25x^2\right)-\left(5x^2-25x\right)+\left(6x-30\right)=0\)

<=> \(x^3\left(x-5\right)+5x^2\left(x-5\right)-5x\left(x-5\right)+6\left(x-5\right)=0\)

<=> \(\left(x-5\right)\left(x^3+5x^2-5x+6\right)=0\)

<=> \(\left(x-5\right)\left[\left(x^3+6x^2\right)-\left(x^2+6x\right)+\left(x+6\right)\right]=0\)

<=> \(\left(x-5\right)\left[x^2\left(x+6\right)-x\left(x+6\right)+\left(x+6\right)\right]=0\)

<=> \(\left(x-5\right)\left(x+6\right)\left(x^2-x+1\right)=0\)

<=> \(\left[{}\begin{matrix}x=5\\x=-6\\x=\varnothing\end{matrix}\right.\)

KL: ...

b) Đặt 2x - 5 = a; x-2 = b

PT <=> \(a^3-b^3=\left(a-b\right)^3\)

<=> \(a^3-b^3=a^3-3a^2b+3ab^2-b^3\)

<=> \(3a^2b-3ab^2=0\)

<=> \(3ab\left(a-b\right)=0\)

TH1: a = 0

<=> 2x - 5 = 0

<=>\(x=\frac{5}{2}\)

Th2: b = 0

<=> x-2 = 0

<=> x = 2

TH3: a - b = 0

<=> 2x - 5 - (x-2) = 0

<=> x = 3

KL: x \(\in\left\{\frac{5}{2};2;3\right\}\)

a) \(8x^3-x=0\)

\(\Leftrightarrow x\left(8x^2-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\8x^2-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt{\frac{1}{8}}\end{cases}}\)

b) \(x\left(x-5\right)=2x-10\)

\(\Leftrightarrow x\left(x-5\right)=2\left(x-5\right)\)

\(\Leftrightarrow x\left(x-5\right)-2\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=5\end{cases}}\)

c) \(3x^2+30x=-75\)

\(\Leftrightarrow x^2+10x=-25\)

\(\Leftrightarrow x^2+10x+25=0\)

\(\Leftrightarrow\left(x+5\right)^2=0\)

\(\Leftrightarrow x+5=0\Leftrightarrow x=-5\)

a: =>5-x=-23

=>x=5+23=28

b: =>x-3-x+7-25+x=54

=>x-21=54

=>x=75

c: =>7-9x-2x+4=-5x-35+27-25=-5x-37

=>-11x+3=-5x-37

=>-6x=-40

=>x=20/3

a) \(10-x-5=\left(-5\right)-7-11\)

\(\Rightarrow5-x=-23\)

\(\Rightarrow x=5+23\)

\(\Rightarrow x=28\)

b) \(\left(x-3\right)-\left(x+17-24\right)-\left(25-x\right)=24-\left(-30\right)\)

\(\Rightarrow x-3-x-17+24-25+x=54\)

\(\Rightarrow x-21=54\)

\(\Rightarrow x=54+21\)

\(\Rightarrow x=75\)

c) \(\left(7-9x\right)-\left(2x-4\right)=-\left(5x+35\right)-\left(-27\right)-25\)

\(\Rightarrow7-9x-2x+4=-5x-35+27-25\)

\(\Rightarrow11-11x=-5x-33\)

\(\Rightarrow-11x+5x=-33-11\)

\(\Rightarrow-6x=-22\)

\(\Rightarrow x=\dfrac{22}{6}=\dfrac{11}{3}\)

a. 
10-x-5 = (-5) - 7 -11
=>5-x = 0
=>x=5
b
(x-3) - (x+17-24) - (25-x) = 24 - (-30)
=>x - 3 - x - 17 + 24 - 25 - x = 24 + 30
=>-x - 21 = 54
=>-x = 75
=>x = -75
c
(7 - 9x) - (2x - 4) = - (5x + 35) - (-27) - 25
=>7-9x - 2x + 4 = -5x - 35 + 27 - 35
=>11 - 11x + 5x = -43
=>16x = 11 + 43
=>16x = 54
=>x=4

Quãng đường đi với vận tốc 30km/h là :

S1=v1.t1=30x (km)

Quãng đường đi với vận tốc 35 km/h là :
S2=v2.t2 = 35y (km)
Tổng quãng đường đi được là
S = S1 + S2 = 30x + 35y

8x²+30x+7=0

 8x²+16x+14x+7=0

8x(x+2) +7(x+2)=0

(8x+7)(x+2)=0

=>\(\orbr{\begin{cases}8x+7=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{7}{8}\\x=-2\end{cases}}}\)

a)

4x²-8x+4=2(1-x)(x+1)

4x²-8x+4-2+2x²=0

6x²-8x+2=0

2(3x²-4x+1)=0

3x²-3x-x+1=0

3x(x-1) -(x-1)=0

(3x-1)(x-1)=0

\(\Rightarrow\orbr{\begin{cases}3x-1=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=1\end{cases}}}\)

d,

x²+3x-18=0

=> x²-3x+6x-18=8

=> x(x-3)+6(x-3)=0

=> (x-3)(x+6)=0

=> \(\orbr{\begin{cases}x=3\\x=-6\end{cases}}\)

\(x^4-30x^2+31x-30=0\)

\(\left(x^4+x\right)-30\left(x^2-x+1\right)=0\)

\(x\left(x^3+1\right)-30\left(x^2-x+1\right)=0\)

\(x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)=0\)

\(\left(x^2-x+1\right)\left[x\left(x+1\right)-30\right]=0\)

Ta có: \(x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\)

\(\Rightarrow x^2+x-30=0\left(x^2-x+1\ne0\right)\)

\(\left(x^2-5x\right)+\left(6x-30\right)=0\)

\(x\left(x-5\right)+6\left(x-5\right)=0\)

\(\left(x-5\right)\left(x+6\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-6\end{cases}}}\)

Vậy \(\orbr{\begin{cases}x=5\\x=-6\end{cases}}\)

Thanks bạn nha