Bài 1:tìm X,biết: a)(4x+1)2 =\(\dfrac{4}{9}\) b)(3x-1)2=25 c)(x-\(\dfrac{1}{3}\))2 =\(\dfrac{1}{4}\) d)(4x-3)2=16
Bài 2: Cho biểu thức B=(\(\dfrac{3X}{2X+3}\)+\(\dfrac{4}{3-2x}\)-\(\dfrac{4x^2-23x-12}{4x^2-9}\)):(\(\dfrac{x+3}{2x+3}\) )với x khác 3/2;-3/2;-3
a) Rút gọn B
b) Tính giá trị của B biết 2x^2+7x+3=0
c) Tìm x thuộc Z để B thuộc Z
d) Tìm x để |B|<1
CỨU MÌNH CÂU d NHA MÌNH CẢM ƠN!
a: \(B=\dfrac{3x\left(2x-3\right)-4\left(2x+3\right)-4x^2+23x+12}{\left(2x-3\right)\left(2x+3\right)}\cdot\dfrac{2x+3}{x+3}\)
\(=\dfrac{6x^2-9x-8x-12-4x^2+23x+12}{2x-3}\cdot\dfrac{1}{x+3}\)
\(=\dfrac{2x^2+6x}{\left(2x-3\right)}\cdot\dfrac{1}{x+3}=\dfrac{2x}{2x-3}\)
b: 2x^2+7x+3=0
=>(2x+3)(x+2)=0
=>x=-3/2(loại) hoặc x=-2(nhận)
Khi x=-2 thì \(A=\dfrac{2\cdot\left(-2\right)}{-2-3}=\dfrac{-4}{-7}=\dfrac{4}{7}\)
d: |B|<1
=>B>-1 và B<1
=>B+1>0 và B-1<0
=>\(\left\{{}\begin{matrix}\dfrac{2x+2x-3}{2x-3}>0\\\dfrac{2x-2x+3}{2x-3}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-3< 0\\\dfrac{4x-3}{2x-3}>0\end{matrix}\right.\Leftrightarrow x< \dfrac{3}{4}\)
Bài 3: Tìm x biết:
a) \(\sqrt{3x-2}=4\)
b) \(\sqrt{4x^2+4x+1}-11=5\)
Bài 4: Cho biểu thức
C= \(1\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\) (x > 0, x ≠ 1)
a) Rút gọn C
b) Tìm x để C - 6 < 0
Helpp!!!
Bài 3:
a) \(\sqrt{3x-2}=4\)
⇔\(\sqrt{3x-2}=\sqrt{4^2}\)
⇔\(3x-2=4^2=16\)
\(3x=16+2=18\)
\(x=18:3=6\)
Vậy \(x=6\)
b)\(\sqrt{4x^2+4x+1}-11=5\)
⇔\(\sqrt{\left(2x\right)^2+2\left(2x\right)\cdot1+1^2}-11=5\)
⇔\(\sqrt{\left(2x+1\right)^2}-11=5\)
TH1:
⇔\(\left(2x+1\right)-11=5\)
\(2x+1=5+11=16\)
\(2x=16-1=15\)
\(x=15:2=7,5\)
TH2:
⇔\(\left(2x+1\right)-11=-5\)
\(2x-1=-5+11=6\)
\(2x=6+1=7\)
\(x=7:2=3,5\)
Vậy \(x=\left\{7,5;3,5\right\}\)
(Câu này mình không chắc chắn lắm)
(Học sinh lớp 6 đang làm bài này)
Bài 4:
a: \(C=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\)
\(=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-\sqrt{x}+x+\sqrt{x}}{\sqrt{x}}=\dfrac{2x}{\sqrt{x}}=2\sqrt{x}\)
b: C-6<0
=>C<6
=>\(2\sqrt{x}< 6\)
=>\(\sqrt{x}< 3\)
=>0<=x<9
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0< x< 9\\x\ne1\end{matrix}\right.\)
Bài 3
a)\(\sqrt{3x-2}=4\Leftrightarrow3x-2=16\Leftrightarrow3x=18\Leftrightarrow x=6\)
Vậy PT có nghiệm x=6
b)\(\sqrt{4x^2+4x+1}-11=5\Leftrightarrow\sqrt{\left(2x+1\right)^2}=16\Leftrightarrow2x+1=16hoặc2x+1=-16\)
+)TH1: \(2x+1=16\Leftrightarrow x=\dfrac{15}{2}\Leftrightarrow x=7,5\)
+)TH2:\(2x+1=-16\Leftrightarrow x=\dfrac{17}{2}\Leftrightarrow x=8,5\)
Bài 4
a)\(C=1\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\Leftrightarrow C=\dfrac{x-1}{\sqrt{x}}\left(\dfrac{x-\sqrt{x}+x+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\Leftrightarrow C=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}}\dfrac{2x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\Leftrightarrow C=\dfrac{2x}{\sqrt{x}}\Leftrightarrow C=2\sqrt{x}\)
\(Vậy\) \(C=2\sqrt{x}\)
Bài 2: Tìm x, biết: a) (x+2)(x² -2x+4)-x(x²+2)=15 b) (x-2)³-(x-4)(x² + 4x+16) + 6(x+1)=49 c) (x - 1)³ + (2 - x)(4 + 2x + x²)+ 3x(x + 2) = 16 d) (x - 3)³ - (x - 3)(x² + 3x + 9) + 9(x + 1)² = 15
a: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow2x=-7\)
hay \(x=-\dfrac{7}{2}\)
b: Ta có: \(\left(x-2\right)^3-\left(x-4\right)\left(x^2+4x+16\right)+6\left(x+1\right)^2=49\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6\left(x+1\right)^2=49\)
\(\Leftrightarrow-6x^2+12x+56+6x^2+12x+6=49\)
\(\Leftrightarrow24x=-13\)
hay \(x=-\dfrac{13}{24}\)
Tìm giá trị nhỏ nhất của biểu thức
a) \(A=\sqrt{x-2}+\sqrt{4-x}\)
b) \(y=\dfrac{4x^4-3x^2+9}{x^2},x\ne0\)
c) \(P=\dfrac{x}{4}+\dfrac{1}{x-1}\) với x>1
\(A=\sqrt{x-2}+\sqrt{4-x}\ge\sqrt{x-2+4-x}=\sqrt{2}\)
\(A_{min}=\sqrt{2}\) khi \(\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
\(y=4x^2+\dfrac{9}{x^2}-3\ge2\sqrt{\dfrac{36x^2}{x^2}}-3=9\)
\(y_{min}=9\) khi \(x^2=\dfrac{3}{2}\)
\(P=\dfrac{x-1}{4}+\dfrac{1}{x-1}+\dfrac{1}{4}\ge2\sqrt{\dfrac{x-1}{4\left(x-1\right)}}+\dfrac{1}{4}=\dfrac{5}{4}\)
\(P_{min}=\dfrac{5}{4}\) khi \(x=\dfrac{3}{2}\)
\(a,\dfrac{1}{x^2+3x+2}-\dfrac{3}{x^2-x-2}=\dfrac{-1}{x^2-4}\)
\(b,\dfrac{2x-1}{x^2+4x-5}+\dfrac{x-2}{x^2-10x+9}=\dfrac{3x-12}{x^2-4x-45}\)
a) ĐKXĐ: \(x\notin\left\{-1;-2;2\right\}\)
Ta có: \(\dfrac{1}{x^2+3x+2}-\dfrac{3}{x^2-x-2}=\dfrac{-1}{x^2-4}\)
\(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+2\right)}-\dfrac{3}{\left(x-2\right)\left(x+1\right)}=\dfrac{-1}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\dfrac{x-2}{\left(x+1\right)\left(x+2\right)\left(x-2\right)}-\dfrac{3\left(x+2\right)}{\left(x+2\right)\left(x+1\right)\left(x-2\right)}=\dfrac{-1\left(x+1\right)}{\left(x+1\right)\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(x-2-3x-6=-x-1\)
\(\Leftrightarrow-2x-8+x+1=0\)
\(\Leftrightarrow-x-7=0\)
\(\Leftrightarrow-x=7\)
hay x=-7(thỏa ĐK)
Vậy: S={-7}
a) ĐKXĐ: x∉{−1;−2;2}x∉{−1;−2;2}
Ta có: ⇔1(x+1)(x+2)−3(x−2)(x+1)=−1(x−2)(x+2)⇔1(x+1)(x+2)−3(x−2)(x+1)=−1(x−2)(x+2)
1x2+3x+2−3x2−x−2=−1x2−41x2+3x+2−3x2−x−2=−1x2−4
⇔x−2(x+1)(x+2)(x−2)−3(x+2)(x+2)(x+1)(x−2)=−1(x+1)(x+1)(x−2)(x+2)⇔x−2(x+1)(x+2)(x−2)−3(x+2)(x+2)(x+1)(x−2)=−1(x+1)(x+1)(x−2)(x+2)
Suy ra: x−2−3x−6=−x−1x−2−3x−6=−x−1
⇔−2x−8+x+1=0⇔−2x−8+x+1=0
⇔−x−7=0⇔−x−7=0
⇔−x=7⇔−x=7
hay x=-7(thỏa ĐK)
Vậy: S={-7}
Đọc tiếp
cho \(\lim\limits_{x\rightarrow2}\left(\dfrac{1}{3x^2-4x-4}+\dfrac{1}{x^2-12x+20}\right)=\dfrac{a}{b}\). tìm a,b biết a/b tối giản
\(\lim\limits_{x\rightarrow2}\left(\dfrac{1}{\left(x-2\right)\left(3x+2\right)}+\dfrac{1}{\left(x-2\right)\left(x-10\right)}\right)=\lim\limits_{x\rightarrow2}\dfrac{1}{\left(x-2\right)}\left(\dfrac{x-10+3x+2}{\left(3x+2\right)\left(x-10\right)}\right)\)
\(=\lim\limits_{x\rightarrow2}\dfrac{4\left(x-2\right)}{\left(x-2\right)\left(3x+2\right)\left(x-10\right)}=\lim\limits_{x\rightarrow2}\dfrac{4}{\left(3x+2\right)\left(x-10\right)}=-\dfrac{1}{16}\)
Bài 1)tìm Min hay Max
a) G=\(\dfrac{2}{x^2+8}\)
b) H=\(\dfrac{-3}{x^2-5x+1}\)
Bài 2) Tìm Min hay Max
a)D=\(\dfrac{2x^2-16x+41}{x^2-8x+22}\)
b)E=\(\dfrac{4x^4-x^2-1}{\left(x^2+1\right)^2}\)
c)G=\(\dfrac{3x^2-12x+10}{x^2-4x+5}\)
1.
\(G=\dfrac{2}{x^2+8}\le\dfrac{2}{8}=\dfrac{1}{4}\)
\(G_{max}=\dfrac{1}{4}\) khi \(x=0\)
\(H=\dfrac{-3}{x^2-5x+1}\) biểu thức này ko có min max
2.
\(D=\dfrac{2x^2-16x+41}{x^2-8x+22}=\dfrac{2\left(x^2-8x+22\right)-3}{x^2-8x+22}=2-\dfrac{3}{\left(x-4\right)^2+6}\ge2-\dfrac{3}{6}=\dfrac{3}{2}\)
\(D_{min}=\dfrac{3}{2}\) khi \(x=4\)
\(E=\dfrac{4x^4-x^2-1}{\left(x^2+1\right)^2}=\dfrac{-\left(x^4+2x^2+1\right)+5x^4+x^2}{\left(x^2+1\right)^2}=-1+\dfrac{5x^4+x^2}{\left(x^2+1\right)^2}\ge-1\)
\(E_{min}=-1\) khi \(x=0\)
\(G=\dfrac{3\left(x^2-4x+5\right)-5}{x^2-4x+5}=3-\dfrac{5}{\left(x-2\right)^2+1}\ge3-\dfrac{5}{1}=-2\)
\(G_{min}=-2\) khi \(x=2\)
Bài 1. Tìm x biết:
a) 4x2 -5= 12x+2
b) x2 - 4x+4= 9+12x+4x2
c) x3 -3x2 = 1-3x
Bài 1. Cho BT A = \(\dfrac{4x+1}{x-1}\) và B = \(\dfrac{3x+1}{x^2-1}\) - \(\dfrac{2x}{x-1}\) + \(\dfrac{3x}{x+1}\)
1) Tìm giá trị biểu thức A tại x = 2
2) Rút gọn biểu thức B
3) Tìm tất cả các giá trị của x để /A.B/ = 4x
1: Khi x=2 thì \(A=\dfrac{4\cdot2+1}{2-1}=9\)
2: \(=\dfrac{3x+1-2x^2-2x+3x^2-3x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{x+1}\)