a) 1/2 x 1/4 + 1/2 x 3/4
b) 4/5 x 1/2 - 1/2 x 1/4
Những câu hỏi liên quan
x^4+ 4 Bx^4 + 64 C x^8+ x^7 + 1 Dx^8 +x^4 + 1 E x^5 + x +1
F x^3 + x^2 + 4 G x^ 4 + 2x^2 - 24 H x^3 - 2x -4 I a^4 +4b^4
\(x^4+4=\left(x^2+2x+2\right)\left(x^2-2x+2\right)\)
\(x^4+2x^2-24=\left(x^2+6\right)\cdot\left(x^2-4\right)=\left(x-2\right)\left(x+2\right)\left(x^2+6\right)\)
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tìm x , biết
a) 17/6- x( x-7/6)= 7/4
b) 3/35 - ( 3/5-x)= 2/7
tìm x thuộc Z , biết
3/4-5/6 < x/12 < 1 -( 2/3-1/4)
tìm x biết
a ) 2x-3=x + 1/2
b) 4x- ( x+ 1/2) = 2x - ( 1/2 - 5 )
Bài 1:
a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)
\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)
\(\Leftrightarrow-12x^2+14x+13=0\)
\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)
b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)
\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)
hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
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Bài 3:
a) Ta có: \(2x-3=x+\dfrac{1}{2}\)
\(\Leftrightarrow2x-x=\dfrac{1}{2}+3\)
\(\Leftrightarrow x=\dfrac{7}{2}\)
b) Ta có: \(4x-\left(x+\dfrac{1}{2}\right)=2x-\left(\dfrac{1}{2}-5\right)\)
\(\Leftrightarrow3x-\dfrac{1}{2}-2x+\dfrac{1}{2}-5=0\)
\(\Leftrightarrow x=5\)
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1/Tính giá trị
a, A=2/3 + 3/4 * (-4/9)
b,B=2 3/11 * 1 1/12 * (-2,2)
c, C= 3/4 - 0,2) : (0,4 - 4/5)
2/ tìm x.
a, 11/12 -x= 2/3 + 1/4
b, 2x (x - 1/7/ = 0
c, 3/4 + 1/4 :x = 2/5
Bài này mình lm r nekk nhưng mak đang kiểm tra lại kết quả , mong các bn iu dấu giải để mik đối chiếu nke ❤ camon mn nhìu <3
1/
a, \(A=\dfrac{2}{3}+\dfrac{3}{4}.\left(-\dfrac{4}{9}\right)=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\)
b, \(B=2\dfrac{3}{11}.\dfrac{11}{12}.\left(-2,2\right)=\dfrac{25}{11}.\dfrac{11}{12}.\left(-\dfrac{11}{5}\right)=-\dfrac{55}{12}\)
c, \(C=\left(\dfrac{3}{4}-0,2\right):\left(0,4-\dfrac{4}{5}\right)=\left(\dfrac{3}{4}-\dfrac{1}{5}\right):\left(\dfrac{2}{5}-\dfrac{4}{5}\right)=\dfrac{11}{20}:\left(-\dfrac{2}{5}\right)=-\dfrac{11}{8}\)
2/
a, \(\dfrac{11}{12}-x=\dfrac{2}{3}+\dfrac{1}{4}\\ \Rightarrow\dfrac{11}{12}-x=\dfrac{11}{12}\\ \Rightarrow x=0\)
b, \(2x\left(x-\dfrac{1}{7}\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)
c, \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\\ \Rightarrow\dfrac{1}{4}:x=-\dfrac{7}{20}\\ \Rightarrow x=-\dfrac{5}{7}\)
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Phân tích thành nhân tử
1, a^4+a^2+1
2,a^4+4b^4
3,64x^4+1
4, x^5+x^4+1
5, x^7+x^2+1
6, x^8+x+1
7, x^4-4x^2+4x-1
8, a^16+a^8b^8+b^16
1)
=a^4+2a^2+1-a^2
=(a^2+1)^2-a^2
=(a^2-a+1)(a^2+a+1)
2)
=a^4+4b^4-4a^2b^2
=(a^2+2b^2)^2-4a^2b^2
=(a^2-2ab+2b^2)(a^2+2ab+2b^2)
3)
=(8x^2+1)^2-16x^2
=(8x^2-4x+1)(8x^2+4x+1).
4)
=x^5+x^4+x^3-x^3+1
=x^2(x^2+x+1)-(x-1)(x^2+x+1)
=(x^2-x+1)(x^2+x+1)
5).
=x^7-x+x^2+x+1
=x(x^6-1)+x^2+x+1
=x(x^3-1)(x^3+1)+x^2+x+1
=x(x-1)(x^2+x+1)(x^3+1)+x^2+x+1
=(x^2+x+1)[(x^2-x)(x^3+1)+1]
6)
=x^8-x^2+x^2+x+1
=x^2(x-1)(x^2+x+1)(x^3+1)+x^2+x+1
Xong nhóm x^2+x+1 vào.
7)
=x^4-(2x-1)^2
=(x^2-2x+1)(x^2+2x-1)
8)
=(a^8+b^8)^2-a^8b^8
=(a^8-a^4b^4+b^8)(a^8+a^4b^4+b^8).
a. x+1/x-2 - x/x+2 + 8/x2 -4
b. x-3/x+1 - x+2/x-1 + 8x/x2 -1
c. x+2/x2-2x + 2/x2+2x + 3x+2/x2-4
d. 4/x - 12/x2+3x + 5/x+3
a: \(=\dfrac{x^2+3x+2-x^2+2x+8}{\left(x-2\right)\left(x+2\right)}=\dfrac{5x+10}{\left(x-2\right)\left(x+2\right)}=\dfrac{5}{x-2}\)
b: \(=\dfrac{x^2-4x+3-x^2-3x-2+8x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x-1}\)
c: \(=\dfrac{x+2}{x\left(x-2\right)}+\dfrac{2}{x\left(x+2\right)}+\dfrac{3x+2}{\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{x^2+2x+2x-4+3x+2}{x\left(x-2\right)\left(x+2\right)}=\dfrac{x^2+7x-2}{x\left(x-2\right)\left(x+2\right)}\)
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a,
\(\dfrac{x+1}{x-2}-\dfrac{x}{x+2}+\dfrac{8}{x^2-4}\\ =\dfrac{x^2+3x+2-x^2+2x+8}{\left(x-2\right)\left(x+2\right)}=\dfrac{5x+10}{\left(x-2\right)\left(x+2\right)}=\dfrac{5\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{5}{x-2}\)
b,
\(\dfrac{x-3}{x+1}-\dfrac{x+2}{x-1}+\dfrac{8x}{x^2-1}\\ =\dfrac{x^2-4x+3-x^2-3x-2+8x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{1}{x-1}\)
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a. |x+ 2/5|- 2= -1/4
b. 1/5 + |x- 13/10| = 3/2
c. |3/4 - 1/2x| + 1/3 = 5/6
d. 7,5 -3 |5- 2x| = -4,5
đ. | x - 3,5| + | x - 1,3| = 0
e. |x- 2021| + | x- 2022| = 0
f. |x| + x = 1/3
g. |x- 2| = x
giúp mik với ạ, mik đang cần gấp
\(a,\Leftrightarrow\left|x+\dfrac{2}{5}\right|=\dfrac{7}{4}\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{2}{5}=\dfrac{7}{4}\left(x\ge-\dfrac{2}{5}\right)\\x+\dfrac{2}{5}=-\dfrac{7}{4}\left(x< -\dfrac{2}{5}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{27}{20}\left(tm\right)\\x=-\dfrac{43}{20}\left(tm\right)\end{matrix}\right.\)
\(b,\Leftrightarrow\left|x-\dfrac{13}{10}\right|=\dfrac{13}{10}\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{13}{10}=\dfrac{13}{10}\left(x\ge\dfrac{13}{10}\right)\\x-\dfrac{13}{10}=-\dfrac{13}{10}\left(x< \dfrac{13}{10}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{5}\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)
\(c,\Leftrightarrow\left|\dfrac{3}{4}-\dfrac{1}{2}x\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}-\dfrac{1}{2}x=\dfrac{1}{2}\left(x\le\dfrac{3}{2}\right)\\\dfrac{1}{2}x-\dfrac{3}{4}=\dfrac{1}{2}\left(x>\dfrac{3}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{5}{2}\left(tm\right)\end{matrix}\right.\)
\(d,\Leftrightarrow\left|5-2x\right|=4\Leftrightarrow\left[{}\begin{matrix}5-2x=4\left(x\le\dfrac{5}{2}\right)\\2x-5=4\left(x>\dfrac{5}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{9}{2}\left(tm\right)\end{matrix}\right.\)
\(đ,\Leftrightarrow\left\{{}\begin{matrix}x-3,5=0\\x-1,3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3,5\\x=1,3\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)
\(e,\Leftrightarrow\left\{{}\begin{matrix}x-2021=0\\x-2022=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2021\\x=2022\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)
\(f,\Leftrightarrow\left|x\right|=\dfrac{1}{3}-x\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}-x\left(x\ge0\right)\\x=x-\dfrac{1}{3}\left(x< 0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\left(tm\right)\\0x=-\dfrac{1}{3}\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{6}\)
\(g,\Leftrightarrow\left[{}\begin{matrix}x-2=x\left(x\ge2\right)\\2-x=x\left(x< 2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0x=2\left(vô.lí\right)\\x=1\left(tm\right)\end{matrix}\right.\Leftrightarrow x=1\)
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phân tích đa thức thành nhân tử
1)ab(a+b)-2bc(b-2c)-2ca(a-2c)-4abc
2)a^2b+2ab^2+4b^2c+4bc^2+2c^2a+ca^2+4abc
3)(x^2-6x+5)(x^2-10x+21)-20
4)4(x^2+x+1)^2+5x(x^2+x+1)+x^2
5)x^4+5x^3-12x^2+5x+1
6)(x+1)(x-4)(x+2)(x-8)+4x^2
7)4x^3+5x^2+10x-12
8)(x+3)^2(3x+8)(3x+10)-8
9)(4x+1)(12x-1)(3x+2)(x+1)-4
tìm x thỏa mãn:
a) (x+2)(x+3)-(x-2)(x-5)=-4
b) (x+1)(x2-x+1)-x(x-3)(x+3)=8
c) 4x2-9=(3x+1)(2x-3)
d) (3x+1)2-4(x-1)2=0
a: Ta có: \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x-5\right)=-4\)
\(\Leftrightarrow x^2+5x+6-x^2+7x-10=-4\)
\(\Leftrightarrow12x=0\)
hay x=0
b: Ta có: \(\left(x+1\right)\left(x^2-x+1\right)-x\left(x-3\right)\left(x+3\right)=8\)
\(\Leftrightarrow x^3+1-x^3+9x=8\)
\(\Leftrightarrow9x=7\)
hay \(x=\dfrac{7}{9}\)
c: Ta có: \(4x^2-9=\left(3x+1\right)\left(2x-3\right)\)
\(\Leftrightarrow\left(3x+1\right)\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(3x+1-2x-3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\end{matrix}\right.\)
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Phân tích thành nhân tử ( thêm bớt hạng tử)
1, a^4+a^2+1
2,a^4+4b^4
3,64x^4+1
4, x^5+x^4+1
5, x^7+x^2+1
6, x^8+x+1
7, x^4-4x^2+4x-1
8, a^16+a^8b^8+b^16
1, a4 + a2 + 1
= a4 + 2a2 + 1 - a2
= (a2)2 + 2a2 + 1 - a2
= (a2 + 1)2 - a2
= (a2 + 1 - a)(a2 + 1 + a)
2, a4 + 4b4
= (a2)2 + 2. a2 . b2 + (2b)2 - a2 . b2
= (a2 + 2b)2 - (ab)2
= (a2 + 2b - ab)(a2 + 2b + ab)
3, 64x4 + 1
= (8x2)2 + 16x2 + 1 - 16x2
= (8x2 + 1)2 - (4x)2
= (8x2 + 1 - 4x)(8x2 + 1 + 4x)
4, x5 + x4 + 1
= x5 + x4 + x3 - x3 - x2 - x + x + x2 + 1
= (x5 + x4 + x3) - (x3 + x2 + x) + (x + x2 + 1)
= x3(x2 + x + 1) - x(x2 + x + 1) + (x2 + x + 1)
= (x2 + x + 1)(x3 - x + 1)
5, x7 + x2 + 1
= x7 – x + x2 + x + 1
= x(x6 – 1) + (x2 + x + 1)
= x(x3 – 1)(x3 + 1) + (x2 + x + 1)
= x(x3 + 1)(x – 1) (x2 + x + 1) + (x2 + x + 1)
= (x2 + x + 1)[ x(x3 + 1)(x – 1) + 1]
= (x2 + x + 1)(x5 – x4 + x3 – x2 + x – 1)
6, x8 + x + 1
= x8 + x7 + x6 - x7 - x6 - x5 + x5 + x4 + x3 - x4 - x3 - x2 + x2 + x + 1
= (x8 + x7 + x6) - (x7 + x6 + x5) + (x5 + x4 + x3 ) - (x4 + x3 + x2) + (x2 + x + 1)
= x6(x2 + x + 1) - x5(x2 + x + 1) + x3(x2 + x + 1) - x2(x2 + x + 1) + (x2 + x + 1)
= (x2 + x + 1)(x6 - x5 + x3 - x2 + 1)
7, x4 - 4x2 + 4x - 1
= x4 - (4x2 - 4x + 1)
= (x2)2 - (2x - 1)2
= (x2 - 2x + 1)(x2 + 2x - 1)
= (x - 1)2 (x2 + 2x - 1)
8, a16 + a8b8 + b16
= (a16 + 2a8b8 + b16) - a8b8
= (a8 + b8)2 - (a4b4)2
= (a8 + b8 - a4b4)(a8 + b8 + a4b4)
= (a8 + b8 - a4b4)[(a8 + b8 + 2a4b4) - a4b4]
= (a8 + b8 - a4b4)[(a4 + b4)2 - (a2b2)2]
= (a8 + b8 - a4b4)(a4 + b4 - a2b2)(a4 + b4 + a2b2)
= (a8 + b8 - a4b4)(a4 + b4 - a2b2)[(a4 + b4 + 2a2b2) - a2b2]
= (a8 + b8 - a4b4)(a4 + b4 - a2b2)[(a2 + b2) - (ab)2]
= (a8 + b8 - a4b4)(a4 + b4 - a2b2)(a2 + b2 - ab)(a2 + b2 + ab)