A=1(x+3) - | 3x+2 |
a)rut gon A
b)tim x de A=5
Mik can rat gap giup mik di mik tick cho :(
Những câu hỏi liên quan
cho M=((x+2/3x)+(2/x+1)-3):(2-4x/x+1)-(3x-3x^2+1/3x)
a. rut gon M
b. tim x sao cho M<1/3
c. tim xϵZ de MϵZ
giup mk vs, mk dang can gap a
cho bieu thuc A=(x-√x/√-1+1):(x+√x/√x+1) (x≥0;x≠1)
a. tim x de bieu thuc A co nghia ? rut gon A?
b. tinh gia tri cua bieu thuc A tai x =7+4√3
lm giup mik nha
căn bậc hai không có số âm
\(\sqrt{-1}\) đó
a) ĐK : x ≥ 0 ; x ≠ 1
A=\(\frac{x-\sqrt{x}}{\sqrt{x}-1}:\frac{x+\sqrt{x}}{\sqrt{x}+1}\)
=\(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}:\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
=\(\sqrt{x}:\sqrt{x}\)
=1
Vậy A=1 với x ≥ 0 ; x ≠ 1
b) Vì A=1 nên không thể thay x
cho 2 bieu thuc A=x+x^2/2-x va B=2x/x+1+3/x-2-2x^2+1/x^2-x-2 a, tinh gia tri cua A khi /2x-3/=1 b,tim dieu kien xac dinh va rut gon bieu thuc B c,tim so nguyen x de P=A.B dat gia tri lon nhat
mk dang can gap
a:
ĐKXĐ: x<>2
|2x-3|=1
=>\(\left[{}\begin{matrix}2x-3=1\\2x-3=-1\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=2\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)
Thay x=1 vào A, ta được:
\(A=\dfrac{1+1^2}{2-1}=\dfrac{2}{1}=2\)
b: ĐKXĐ: \(x\notin\left\{-1;2\right\}\)
\(B=\dfrac{2x}{x+1}+\dfrac{3}{x-2}-\dfrac{2x^2+1}{x^2-x-2}\)
\(=\dfrac{2x}{x+1}+\dfrac{3}{x-2}-\dfrac{2x^2+1}{\left(x-2\right)\left(x+1\right)}\)
\(=\dfrac{2x\left(x-2\right)+3\left(x+1\right)-2x^2-1}{\left(x+1\right)\left(x-2\right)}\)
\(=\dfrac{2x^2-4x+3x+3-2x^2-1}{\left(x+1\right)\left(x-2\right)}\)
\(=\dfrac{-x+2}{\left(x+1\right)\left(x-2\right)}=-\dfrac{1}{x+1}\)
c: \(P=A\cdot B=\dfrac{-1}{x+1}\cdot\dfrac{x\left(x+1\right)}{2-x}=\dfrac{x}{x-2}\)
\(=\dfrac{x-2+2}{x-2}=1+\dfrac{2}{x-2}\)
Để P lớn nhất thì \(\dfrac{2}{x-2}\) max
=>x-2=1
=>x=3(nhận)
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cho bieu thuc P= (\(\frac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}+\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}-3}\) ): \(\frac{1}{x-1}\)
a) Tim dieu kien de P co nghia, rut gon bieu thuc P.
b) Tim cac so tu nhien x de \(\frac{1}{P}\)la so tu nhien
c) Tinh gia tri cua P voi x= 4-\(2\sqrt{3}\)
Giup mk vs mk dang can gap
Cho M=\(\frac{a^4-16}{a^4-4a^3+8a^2-16a+16}\)
Rut gon roi tim gia tri nguyen cua a de M nguyen.
Giup minh nhe minh dang can gap.
M = \(\frac{a^4-16}{a^4-4a^3+8a^2-16a+16}\)
=> M = \(\frac{\left(a^2+4\right)\left(a^2-4\right)}{\left(a^4-4a^3+4a^2\right)+\left(4a^2-16a+16\right)}\)
M = \(\frac{\left(a-2\right)\left(a+2\right)\left(a^2+4\right)}{a^2\left(a^2-4a+4\right)+4\left(a^2-4a+4\right)}\)
M = \(\frac{\left(a-2\right)\left(a+2\right)\left(a^2+4\right)}{\left(a^2+4\right)\left(a^2-4a+4\right)}\)
M = \(\frac{\left(a-2\right)\left(a+2\right)\left(a^2+4\right)}{\left(a^2+4\right)\left(a-2\right)^2}\)
M = \(\frac{a+2}{a-2}\)
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Cho bieu thuc:
\(P=\frac{x+2}{x+3}-\frac{5}{x^2+x+6}+\frac{1}{2-x}\)voi \(x\ne-3;x\ne2\)
a, Rut gon P
b, tim x de \(P=\frac{-3}{4}\)
c,tim x nguyen de P dat gia tri nguyen
(mk dang can gap ai giai dung dau tien mk tick cho)
f(x) = 2 (x^3 - x) + x^2 - 3x + 3 - x^3
a) Thu gon hai da thuc tren roi xep theo luy thua giam dan cua bien
b) Tinh f(-1);g (1/2)
c) Tim f(x) + g(x);f(x) - g(x)
d) Tim gia tri cua x de f(x) = g(x) (giup mk voi mk dang can gap)
Bai 1: A= \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\) B= \(\left(\dfrac{\sqrt{X}+1}{\sqrt{X}-1}-\dfrac{\sqrt{X}-1}{\sqrt{X}+1}\right)\) : \(\dfrac{\sqrt{X}}{\sqrt{X}-1}\) ( X> 0, X≠1)
A) Rut B
b) Tim x de gia tri cua A va B trai dau
(mink dag can rat gap)
a) Ta có: \(B=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
\(=\left(\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right):\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
\(=\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
\(=\dfrac{4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
\(=\dfrac{4}{\sqrt{x}+1}\)
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b. Để A và B trái dấu \(\Leftrightarrow AB< 0\)
\(\Leftrightarrow\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right)\left(\dfrac{4}{\sqrt{x}+1}\right)< 0\)
\(\Leftrightarrow\dfrac{4}{\sqrt{x}-1}< 0\Leftrightarrow\sqrt{x}-1< 0\)
\(\Rightarrow0< x< 1\)
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cho biểu thức A left(frac{3}{4}+frac{3}{x^2+3x}right)chialeft(frac{x^2}{27-3x}+frac{1}{x+3}right)a, rut gon A b, tim x de a-1c, tim cac gia tri nguyen cua x de A co gia tri nguyencho bthuc B left(frac{x^2}{x^3-4x}+frac{6}{6-3x}+frac{1}{x-2}right)chialeft(x-2+frac{16-x^2}{x+2}right)rut gon B tính b khi /x/ 1/2tim x de b2tim x in z de b in z
Đọc tiếp
cho biểu thức A= \(\left(\frac{3}{4}+\frac{3}{x^2+3x}\right)chia\left(\frac{x^2}{27-3x}+\frac{1}{x+3}\right)\)
a, rut gon A
b, tim x de a<-1
c, tim cac gia tri nguyen cua x de A co gia tri nguyen
cho bthuc B = \(\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x-2}\right)chia\left(x-2+\frac{16-x^2}{x+2}\right)\)rut gon B tính b khi /x/ = 1/2tim x de b=2tim x \(\in\) z de b \(\in\) z
Bài 2:
a: \(B=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{6}{3\left(x-2\right)}+\dfrac{1}{x-2}\right):\left(\dfrac{x^2-4+16-x^2}{x+2}\right)\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)
\(=\dfrac{x-x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{12}=\dfrac{-1}{6\left(x-2\right)}\)
b: Thay x=1/2 vào B, ta được:
\(B=\dfrac{-1}{6\cdot\left(\dfrac{1}{2}-2\right)}=\dfrac{-1}{6\cdot\dfrac{-3}{2}}=\dfrac{1}{9}\)
Thay x=-1/2 vào B, ta được:
\(B=\dfrac{-1}{6\cdot\left(-\dfrac{1}{2}-2\right)}=-\dfrac{1}{15}\)
c: Để B=2 thì \(\dfrac{-1}{6\left(x-2\right)}=2\)
=>6(x-2)=-1/2
=>x-2=-1/12
hay x=23/12
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