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binh rồi căn thì cứ chuyển bỏ dấu âm đi nó tương tự dấu giá trị tuyệt đối thôi
a, \(x^2-9=0\Rightarrow x^2=9\Rightarrow x\pm3\)
b, \(\left(x-3\right)^2-25=0\Rightarrow\left(x-3\right)^2=25\)
\(\Rightarrow\left\{{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
c, \(\left(x-3\right)\left(2x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\2x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\2x=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{5}{2}\end{matrix}\right.\)
d, \(\left(x-3\right)x-2\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
e, \(3x\left(x-1\right)-5\left(1-x\right)=0\)
\(\Rightarrow3x\left(x-1\right)+5\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(3x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\3x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{3}\end{matrix}\right.\)
g, \(x^2+6x-7=0\)
\(\Rightarrow x^2-x+7x-7=0\)
\(\Rightarrow x.\left(x-1\right)+7.\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x+7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
h,\(2x^2+5x-7=0\)
\(\Rightarrow2x^2-2x+7x-7=0\)
\(\Rightarrow2x.\left(x-1\right)+7.\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(2x+7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\)
Chúc bạn học tốt!!!
a) \(x^2-9=0\Leftrightarrow x^2=9\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\) vậy \(x=3;x=-3\)
b) \(\left(x-3\right)^2-25=0\Leftrightarrow\left(x-3\right)^2=25\Leftrightarrow\left\{{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
vậy \(x=8;x=-2\)
c) \(\left(x-3\right)\left(2x-5\right)=0\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\2x-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=\dfrac{5}{2}\end{matrix}\right.\)
vậy \(x=3;x=\dfrac{5}{2}\)
d)\(\left(x-3\right).x-2\left(x-3\right)=0\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=3\end{matrix}\right.\) vậy \(x=2;x=3\)
e) \(3x\left(x-1\right)-5\left(1-x\right)=0\Leftrightarrow\left(3x+5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+5=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-5}{3}\\x=1\end{matrix}\right.\) vậy \(x=\dfrac{-5}{3};x=1\)
câu e t thấy sai sai nhưng vẫn làm ; bn coi lại đề nha
g) \(x^2+6x-7=0\Leftrightarrow x^2-x+7x-7=0\)
\(\Leftrightarrow x\left(x-1\right)+7\left(x-1\right)=0\Leftrightarrow\left(x+7\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+7=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-7\\x=1\end{matrix}\right.\) vậy \(x=-7;x=1\)
h) \(2x^2+5x-7=0\Leftrightarrow2x^2-2x+7x-7=0\)
\(\Leftrightarrow2x\left(x-1\right)+7\left(x-1\right)=0\Leftrightarrow\left(2x+7\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+7=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-7}{2}\\x=1\end{matrix}\right.\) vậy \(x=\dfrac{-7}{2};x=1\)
\(c,-5\left(2-x\right)+4\left(x-3\right)=10x-15\)
\(-10+5x+4x-12=10x-15\)
\(5x+4x-10x-10-12=-15\)
\(-x-10=-3\)
\(-x=7\)
\(x=-7\)
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
b/100x+(1+2+3+...+100)=205550
100x+5050=205550
100x=205550-5050
100x=200500
x=200500/100
x=2005
d/(3x-24).75=2.76.1/20090
(3x-24).75=2.76.1
(3x-24)=2.76:75
(3x-24)=2.7
3x-16=14
3x=14+16
3x=30
x=30/10=3
Trả lời
Mk nghĩ bạn có thể tham khảo ở CHTT nha !
Có đáp án của câu b;c và d đó.
Đừng ném đá chọi gạch nha !
a) vi(x^2+5)(x^2-25)=0
=>x^2+5=0 hoac x^2-25=0
=>x=...hoac x=...(tu lam)
b)(x-2)(x+1)=0
=>x-2=0 hoac x+1=0
=>x=2 hoac x=-1
c)(x^2+7)(x^2-49)<0
=>x^2+7va x^2-49 trai dau
ma x^2+7>=7=>x^2-49<0=>x<7 va x>-7
con lai tuong tu
tu lam nhe nho k nha