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\(3.\left(x-2\right)+1=25\Leftrightarrow\left(x-2\right)=8\Leftrightarrow x=10\)
a)\(\left(2x-3\right)\left(x+1\right)< 0\)
\(\Leftrightarrow\begin{cases}2x-3>0\\x+1< 0\end{cases}\) hoặc \(\begin{cases}2x-3< 0\\x+1>0\end{cases}\)
\(\Leftrightarrow\begin{cases}x>\frac{3}{2}\\x< -1\end{cases}\) (loại) hoặc \(\begin{cases}x< \frac{3}{2}\\x>-1\end{cases}\)
\(\Leftrightarrow-1< x< \frac{3}{2}\)
b) \(\left(x-\frac{1}{2}\right)\left(x+3\right)>0\)
\(\Leftrightarrow\begin{cases}x-\frac{1}{2}>0\\x+3>0\end{cases}\) hoặc \(\begin{cases}x-\frac{1}{2}< 0\\x+3< 0\end{cases}\)
\(\Leftrightarrow\begin{cases}x>\frac{1}{2}\\x>-3\end{cases}\) hoặc \(\begin{cases}x< \frac{1}{2}\\x< -3\end{cases}\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x>\frac{1}{2}\\x< -3\end{array}\right.\)
c) Sai đề phải là \(\frac{x}{\left(x+3\right)\left(x+7\right)}\)
Có: \(\frac{3}{\left(x+3\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+3\right)\left(x+17\right)}\)
\(\Leftrightarrow\)\(\frac{1}{x+3}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+7}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)
\(\Leftrightarrow\)\(\frac{1}{x+3}-\frac{1}{x+7}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)
\(\Leftrightarrow\)\(\frac{4}{\left(x+3\right)\left(x+7\right)}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)
\(\Leftrightarrow x=4\)
\(8,1-\left(x-6\right)=4\left(2-2x\right)\)
\(\Leftrightarrow1-x+6=8-8x\)
\(\Leftrightarrow-x+8x=8-1-6\)
\(\Leftrightarrow7x=1\)
\(\Leftrightarrow x=\dfrac{1}{7}\)
\(9,\left(3x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)
\(10,\left(x+3\right)\left(x^2+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\varnothing\end{matrix}\right.\)
`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):}`
\(c,-5\left(2-x\right)+4\left(x-3\right)=10x-15\)
\(-10+5x+4x-12=10x-15\)
\(5x+4x-10x-10-12=-15\)
\(-x-10=-3\)
\(-x=7\)
\(x=-7\)
Vậy \(x=-7.\)
Những câu khác bạn nên tự làm...
a) −7. (5 - x) − 2. (x − 10) = 15
-35 + 7x - 2x + 20 = 15
7x - 2x = 35 - 20 + 15
5x = 30
x = 30 : 5
x = 6
Vậy x = 6
b) 4. (2 − x) + 3(x − 5) = 14
8 - 4x + 3x - 15 = 14
8 - 15 - 14 = 4x - 3x
-21 = x
Vậy x = -21
c) −5. (2 − x) + 4. (x − 3) = 10x −15
-10 + 5x + 4x - 12 = 10x - 15
-10 - 12 + 15 = 10x - 5x - 4x
-7 = x
Vậy x = -7
d) −7. (3x − 5) + 2. (7x − 14) = 28
-21x + 35 + 14x - 28 = 28
35 - 28 - 28 = 21x - 14x
-21 = 7x
x = (-21) : 7
x = -3
Vậy x = -3
e) 5. (4 − x) − 7(−x + 2) = 4 − 9 + 3
20 - 5x + 7x - 14 = -2
-5x + 7x = -2 - 20 + 14
2x = -8
x = (-8) : 2
x = -4
Vậy x = -4
f) 5. (x − 7) + 10. (3 − x) = 20
5x - 35 + 30 - 10x = 20
-35 + 30 - 20 = -5x + 10x
-25 = 5x
x = (-25) : 5
x = -5
Vậy x = -5
g) −4. (x + 1) + 8. (x − 3) = 24
-4x - 4 + 8x - 24 = 24
-4x + 8x = 24 + 4 + 24
4x = 52
x = 52 : 4
x = 13
Vậy x = 13
h) 4. (x − 1) − 3. (x − 2) = −|−5|
4x - 4 - 3x + 6 = -5
4x - 3x = -5 + 4 - 6
x = -7
Vậy x = -7
ta có : \(5^x.5^{\left(x+1\right)}.5^{\left(x+2\right)}\le\dfrac{100...0}{2^{18}}\) \(\Leftrightarrow5^{3x+3}\le\dfrac{10^{18}}{2^{18}}=5^{18}\)
\(\)\(\Leftrightarrow3x+3\le18\Leftrightarrow x\le5\) vậy \(x\le5\)
mình làm câu cuối thôi nhé , những câu còn lại bạn tự làm đi , dễ mà :)))) chỉ cần quy đồng mẫu lên là được
\(=\frac{x+1}{58}+1+\frac{x+2}{57}+1=\frac{x+3}{56}+1+\frac{x+4}{55}\)
\(=\frac{x+59}{58}+\frac{x+59}{57}=\frac{x+59}{56}+\frac{x+59}{55}\)
\(=\frac{x+59}{58}+\frac{x+59}{57}-\frac{x+59}{56}-\frac{x+59}{55}=0\)
\(=\left(x+59\right)\left(\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}\right)=0\)
Vì \(\left(\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}\right)\) luôn khác 0
<=> x + 59 = 0
<=> x=-59
Bài 1:
a) (2x-3). (x+1) < 0
=>2x-3 và x+1 ngược dấu
Mà 2x-3<x+1 với mọi x
\(\Rightarrow\begin{cases}2x-3< 0\\x+1>0\end{cases}\)
\(\Rightarrow\begin{cases}x< \frac{3}{2}\\x>-1\end{cases}\)\(\Rightarrow-1< x< \frac{3}{2}\)
b)\(\left(x-\frac{1}{2}\right)\left(x+3\right)>0\)
\(\Rightarrow x-\frac{1}{2}\) và x+3 cùng dấu
Xét \(\begin{cases}x-\frac{1}{2}>0\\x+3>0\end{cases}\)\(\Rightarrow\begin{cases}x>\frac{1}{2}\\x>-3\end{cases}\)
Xét \(\begin{cases}x-\frac{1}{2}< 0\\x+3< 0\end{cases}\)\(\Rightarrow\begin{cases}x< \frac{1}{2}\\x< -3\end{cases}\)
=>....
Bài 2:
\(S=\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{999.1001}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{999}-\frac{1}{1001}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{1001}\right)\)
\(=\frac{1}{2}\cdot\frac{998}{3003}\)
\(=\frac{499}{3003}\)
1: Ta có: \(4x^2-36=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
2: Ta có: \(\left(x-1\right)^2+x\left(4-x\right)=11\)
\(\Leftrightarrow x^2-2x+1+4x-x^2=11\)
\(\Leftrightarrow2x=10\)
hay x=5
Ta có :
\(3.\left(x-2\right)+10^0=5^{100}:5^{58}\)
=> 3 . ( x - 2 ) + 1 = 5100 - 58
=> 3 . ( x - 2 ) + 1 = 542
=> 3.x - 6 + 1 = 542
=> 3.x - 5 = 542
=> 3.x = 542 + 5 = 5.( 541 + 1 )
=> x = \(\frac{5.\left(5^{41}+1\right)}{3}\)
Không tìm được kết quả cụ thể nhak...
3.(x-2)+10^0 =5^100:5^58
3.(x-2)+1=5^42
3x-6+1=5^42
3x-5=5^42
3x=5^42+5
3x=5.(5^41+1)
x=5.(5^41+1)/3
mình cá luôn ( sr mình gi hơi chậm)
Ok@@