1⇔ (x – 2)[(x + 2) + (3 – 2x)] = 0
2⇔ (x – 2)(5 – x) = 0
giai thich cach suy tu 1 sang 2
Những câu hỏi liên quan
a ) 32^ -x .16^x =2048 =>x=?
b ) (x-y)^2 +|2x-1| = 0 =>x=? ;y=?
c ) (x-2y)^2 +(y+1)^6 =0 =>x=? ;y=?
d ) (2x-5)^2000 + (3y+4)^2002 < hay = 0
=>x=? ;y=?
giai thich ra nha
25 tháng 1 2021 lúc 21:36
Giai phường trình sau:a, 3x^2+2x-10 e, 4x^2-12x+50 i,2x^2+5x-30b,x^2-5x+60 f, 2x^2+5x+30 j,x^2+6x-160c,x^2-3x+20 g,x^2+x-20d,2x^2-6x+10 h, x^2-4x+30
Đọc tiếp
Giai phường trình sau:
a, \(3x^2+2x-1=0\) e, \(4x^2-12x+5=0\) i,\(2x^2+5x-3=0\)
b,\(x^2-5x+6=0\) f, \(2x^2+5x+3=0\) j,\(x^2+6x-16=0\)
c,\(x^2-3x+2=0\) g,\(x^2+x-2=0\)
d,\(2x^2-6x+1=0\) h, \(x^2-4x+3=0\)
a) Ta có: \(3x^2+2x-1=0\)
\(\Leftrightarrow3x^2+3x-x-1=0\)
\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{-1;\dfrac{1}{3}\right\}\)
b) Ta có: \(x^2-5x+6=0\)
\(\Leftrightarrow x^2-2x-3x+6=0\)
\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Vậy: S={2;3}
c) Ta có: \(x^2-3x+2=0\)
\(\Leftrightarrow x^2-x-2x+2=0\)
\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy: S={1;2}
d) Ta có: \(2x^2-6x+1=0\)
\(\Leftrightarrow2\left(x^2-3x+\dfrac{1}{3}\right)=0\)
mà \(2\ne0\)
nên \(x^2-3x+\dfrac{1}{3}=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{23}{12}=0\)
\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=\dfrac{23}{12}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=\dfrac{\sqrt{69}}{6}\\x-\dfrac{3}{2}=\dfrac{-\sqrt{69}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9+\sqrt{69}}{6}\\x=\dfrac{9-\sqrt{69}}{6}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{9+\sqrt{69}}{6};\dfrac{9-\sqrt{69}}{6}\right\}\)
e) Ta có: \(4x^2-12x+5=0\)
\(\Leftrightarrow4x^2-10x-2x+5=0\)
\(\Leftrightarrow2x\left(2x-5\right)-\left(2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{5}{2};\dfrac{1}{2}\right\}\)
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tim x biet
a) 4(x+1)^2-9(x+2)^2=0
b) 4(x+1)^2+(2x-1)^2-8(x+1)(x-1)=11
giai cach lam giup minh nha
Giai cac bất phương trình:a) (x-3)(x-2)0b) (x+3)(x+4)(x2+2)≥≥ 0c) x−1x−2x−1x−2 ≥≥0d)x+32−xx+32−x≥≥ 0e) (x-3)(x-2)(x+1)0g) 2x−12x−10k) x2 +3x+20m) x2+10Bài 2 ) Giai phương trìnha)/x-1/-3b) /2x+1/0c)/3-2x/4d)/x+1/+3x4e) /x+1-4x/5g) /x-1/+/x-3/2k)/x-1/+/x-2/+/x-3/2
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Giai cac bất phương trình:
a) (x-3)(x-2)<0
b) (x+3)(x+4)(x2+2)≥≥ 0
c) x−1x−2x−1x−2 ≥≥0
d)x+32−xx+32−x≥≥ 0
e) (x-3)(x-2)(x+1)<0
g) 2x−12x−1<0
k) x2 +3x+2>0
m) x2+1<0
Bài 2 ) Giai phương trình
a)/x-1/=-3
b) /2x+1/=0
c)/3-2x/=4
d)/x+1/+3x=4
e) /x+1-4x/=5
g) /x-1/+/x-3/=2
k)/x-1/+/x-2/+/x-3/=2
(x+4/x^2-3x+2)+(x+1/x^2-4x+3)=2x+5/x^2-4x+3
ghi ca cach giai
1)2x^2-4xy+2y^2-32
2)5x^2-5y^2-x+y
3)x^2-9xy+20y^2
4)4x^2(x+y)-x-y
5)16+ty^2+6xt-9t+x^2
Giai bang phuong phap phan tich thanh nhan tu chung bang cach nhom hang tu.
Mai mình nộp bài,mong giaup đó
1. =2(x2-2xy+y2-16)
=2((x-y)2-42)
=2(x-y-4)(x-y+4)
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2. =5(x2-2.1/10.x+1/100-y2+2.1/10y-1/100)
=5.((X-1/10)2-(Y-1/10)2)
=5(X-1/10-Y+1/10)(X-1/10+Y-1/10)
=5(X-Y)(X+Y-2/10)
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3. =(X2-9XY+81/4Y2)-1/4Y2
=(X-9/2Y)^2-1/4Y^2
=(X-9/2Y-1/2Y)(X-9/2Y+1/2Y)
(X-5Y)(X-4Y)
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Xem thêm câu trả lời
giai cac phuong trinh
a)\(2x^4+5x^3+x^2+5x+2=0\)
b)\(\sqrt{x-1}-\sqrt[3]{2-x}=1\)
c)\(x-\sqrt{x}+1=\sqrt{2x^2-30x+2}\)
d)\(2x^2+3x+7=\left(x-5\right)\sqrt{2x^2+1}\)
e)\(\sqrt{x-2}+\sqrt{4-x}=2x^2-5x-1\)
Giai cac phuong trinh sau
1) (5x-4).(4x+6) 0
2) (4x-10).(24+5x) 0
3) (x-3).(2x+1) 0
4) (2x+1).(x2+2) 0
5) (x2+4).(7x-3) 0
6) (x-5).(3-2x).(3x+4) 0
7) (x-2).(3x+5) (2x-4).(x+1)
8) (2x+5).(x-4) (x-5).(4-x)
9) 9x2-1 (3x+1).(2x-3)
10) (2x-1)2 49
11) (5x-3)2 - (4x-7)2 0
12) (2x+7)2 9.(x+2)2
Giúp mình giải gấp .Cám ơn
Đọc tiếp
Giai cac phuong trinh sau
1) (5x-4).(4x+6) = 0
2) (4x-10).(24+5x) = 0
3) (x-3).(2x+1) = 0
4) (2x+1).(x2+2) = 0
5) (x2+4).(7x-3) = 0
6) (x-5).(3-2x).(3x+4) = 0
7) (x-2).(3x+5) = (2x-4).(x+1)
8) (2x+5).(x-4) = (x-5).(4-x)
9) 9x2-1 = (3x+1).(2x-3)
10) (2x-1)2 = 49
11) (5x-3)2 - (4x-7)2 = 0
12) (2x+7)2 = 9.(x+2)2
Giúp mình giải gấp .Cám ơn
1) \(\left(5x-4\right)\left(4x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-4=0\\4x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=4\\4x=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{\dfrac{4}{5};\dfrac{3}{2}\right\}\)
2) \(\left(4x-10\right)\left(24+5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-10=0\\24+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=10\\5x=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-24}{5}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{\dfrac{5}{2};\dfrac{-24}{5}\right\}\)
3) \(\left(x-3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{-1}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{3;\dfrac{-1}{2}\right\}\)
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1. x2+y2-2xy -2x+2=0
2.(x+2)4+(x+8)4=272
3. (x-2)6+(x-4)6=64
giai pt . ghi cach giai gium minh. minh tick cho