giúp mình với ahuhuuu

1) ĐKXĐ: \(x\ge\dfrac{3}{2}\)

2) ĐKXĐ: \(x\le\dfrac{3}{2}\)

3) ĐKXĐ: \(x\le-2\)

4) ĐKXĐ: \(x< \dfrac{1}{4}\)

5) ĐKXĐ: \(x\le-\dfrac{5}{3}\)

1) ĐKXĐ: \(\left[{}\begin{matrix}x\ge2\\x\le1\end{matrix}\right.\)

2) ĐKXĐ: \(\dfrac{x-6}{x-2}\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2< 0\\x-6\ge0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< 2\\x\ge6\end{matrix}\right.\)

3) ĐKXĐ: \(\dfrac{2x-4}{5-x}\ge0\)

\(\Leftrightarrow\dfrac{x-2}{x-5}\le0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2\ge0\\x-5< 0\end{matrix}\right.\Leftrightarrow2\le x< 5\)

GIÚP VỚI Ạ

a: ĐKXĐ: \(x\ge1\)

b: ĐKXĐ: \(x< 0\)

c: ĐKXĐ: \(\left[{}\begin{matrix}x\ge11\\x\le3\end{matrix}\right.\)

1) ĐKXĐ: \(\left\{{}\begin{matrix}2x+11\ge0\\x-1\ge0\end{matrix}\right.\)\(\Leftrightarrow x\ge1\)

2) ĐKXĐ: \(\left\{{}\begin{matrix}-5x\ge0\\x\ne0\end{matrix}\right.\)\(\Leftrightarrow x< 0\)

3) ĐKXĐ: \(7x^2+1\ge0\left(đúng\forall x\right)\Leftrightarrow x\in R\)

4) ĐKXĐ: \(x^2-14x+33\ge0\Leftrightarrow\left(x-11\right)\left(x-3\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-11\ge0\\x-3\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-11\le0\\x-3\le0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x\ge11\\x\le3\end{matrix}\right.\)

5) ĐKXĐ: 

+) \(-x^2+6x+16\ge0\)

\(\Leftrightarrow-\left(x^2-6x+9\right)+25\ge0\)

\(\Leftrightarrow\left(x-3\right)^2\le25\Leftrightarrow-5\le x-3\le5\)

\(\Leftrightarrow-2\le x\le8\)

+) \(3x^2\ne0\Leftrightarrow x\ne0\)

\(\Rightarrow\left\{{}\begin{matrix}-2\le x\le8\\x\ne0\end{matrix}\right.\)

1.

\(x^4-6x^2-12x-8=0\)

\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)

\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)

\(\Leftrightarrow x=1\pm\sqrt{5}\)

3.

ĐK: \(x\ge-9\)

\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)

\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)

Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)

\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)

\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)

\(\Leftrightarrow...\)

2.

ĐK: \(x\ne\dfrac{2\pm\sqrt{2}}{2};x\ne\dfrac{-2\pm\sqrt{2}}{2}\)

\(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{1}{2x+\dfrac{1}{x}+4}+\dfrac{1}{2x+\dfrac{1}{x}-4}=\dfrac{3}{5}\)

Đặt \(2x+\dfrac{1}{x}+4=a;2x+\dfrac{1}{x}-4=b\left(a,b\ne0\right)\)

\(pt\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{5}\left(1\right)\)

Lại có \(a-b=8\Rightarrow a=b+8\), khi đó:

\(\left(1\right)\Leftrightarrow\dfrac{1}{b+8}+\dfrac{1}{b}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{2b+8}{\left(b+8\right)b}=\dfrac{3}{5}\)

\(\Leftrightarrow10b+40=3\left(b+8\right)b\)

\(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=-\dfrac{20}{3}\end{matrix}\right.\)

TH1: \(b=2\Leftrightarrow...\)

TH2: \(b=-\dfrac{20}{3}\Leftrightarrow...\)

Để giải các phương trình này, chúng ta sẽ làm từng bước như sau: 1. 13x(7-x) = 26: Mở ngoặc và rút gọn: 91x - 13x^2 = 26 Chuyển về dạng bậc hai: 13x^2 - 91x + 26 = 0 Giải phương trình bậc hai này để tìm giá trị của x. 2. (4x-18)/3 = 2: Nhân cả hai vế của phương trình với 3 để loại bỏ mẫu số: 4x - 18 = 6 Cộng thêm 18 vào cả hai vế: 4x = 24 Chia cả hai vế cho 4: x = 6 3. 2xx + 98x2022 = 98x2023: Rút gọn các thành phần: 2x^2 + 98x^2022 = 98x^2023 Chia cả hai vế cho 2x^2022: x + 49 = 49x Chuyển các thành phần chứa x về cùng một vế: 49x - x = 49 Rút gọn: 48x = 49 Chia cả hai vế cho 48: x = 49/48 4. (x+1) + (x+3) + (x+5) + ... + (x+101): Đây là một dãy số hình học có công sai d = 2 (do mỗi số tiếp theo cách nhau 2 đơn vị). Số phần tử trong dãy là n = 101/2 + 1 = 51. Áp dụng công thức tổng của dãy số hình học: S = (n/2)(a + l), trong đó a là số đầu tiên, l là số cuối cùng. S = (51/2)(x + (x + 2(51-1))) = (51/2)(x + (x + 100)) = (51/2)(2x + 100) = 51(x + 50) Vậy, kết quả của các phương trình là: 1. x = giá trị tìm được từ phương trình bậc hai. 2. x = 6 3. x = 49/48 4. S = 51(x + 50)

nhầm

a) Ta có: \(A=3\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}+30\)

\(=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}+30\)

\(=14\sqrt{2x}+30\)

b) Ta có: \(B=4\sqrt{\dfrac{25x}{4}}-\dfrac{8}{3}\sqrt{\dfrac{9x}{4}}-\dfrac{4}{3x}\cdot\sqrt{\dfrac{9x^3}{64}}\)

\(=4\cdot\dfrac{5\sqrt{x}}{2}-\dfrac{8}{3}\cdot\dfrac{3\sqrt{x}}{2}-\dfrac{4}{3x}\cdot\dfrac{3x\sqrt{x}}{8}\)

\(=10\sqrt{x}-4\sqrt{x}-\dfrac{1}{2}\sqrt{x}\)

\(=\dfrac{11}{2}\sqrt{x}\)

c) Ta có: \(\dfrac{y}{2}+\dfrac{3}{4}\sqrt{9y^2-6y+1}-\dfrac{3}{2}\)

\(=\dfrac{1}{2}y+\dfrac{3}{4}\left(1-3y\right)-\dfrac{3}{2}\)

\(=\dfrac{1}{2}y+\dfrac{3}{4}-\dfrac{9}{4}y-\dfrac{3}{2}\)

\(=-\dfrac{7}{4}y-\dfrac{3}{4}\)

a) \(P=\dfrac{x^2+3x}{x^2-8x+16}:\left(\dfrac{x+4}{x}+\dfrac{1}{x-4}+\dfrac{19-x^2}{x^2-4x}\right)\left(x\ne0,x\ne4\right)\)

\(=\dfrac{x^2+3x}{\left(x-4\right)^2}:\left(\dfrac{x+4}{x}+\dfrac{1}{x-4}+\dfrac{19-x^2}{x\left(x-4\right)}\right)\)

\(=\dfrac{x^2+3x}{\left(x-4\right)^2}:\dfrac{\left(x+4\right)\left(x-4\right)+x+19-x^2}{x\left(x-4\right)}\)

\(=\dfrac{x^2+3x}{\left(x-4\right)^2}:\dfrac{x+3}{x\left(x-4\right)}=\dfrac{x\left(x+3\right)}{\left(x-4\right)^2}.\dfrac{x\left(x-4\right)}{x+3}=\dfrac{x^2}{x-4}\)

b) \(x=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\)

\(=\sqrt{3}+1-\sqrt{3}+1=2\)

\(\Rightarrow P=\dfrac{2^2}{2-4}=-2\)

a)\(ĐKXĐ:\left\{{}\begin{matrix}x\left(x-4\right)\ne0\\\dfrac{x+4}{x}+\dfrac{1}{x-4}+\dfrac{19-x^2}{x^2-4x}\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne4\\x\ne0\\x\ne-3\end{matrix}\right.\)

\(P=\dfrac{x\left(x+3\right)}{\left(x-4\right)}:\left(\dfrac{x^2-16+x+19-x^2}{x\left(x-4\right)}\right)=\dfrac{x\left(x+3\right)}{\left(x-4\right)^2}.\left(\dfrac{x\left(x-4\right)}{x+3}\right)=\dfrac{x^2}{x-4}\)

b)\(x=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3+1}-\left(\sqrt{3}-1\right)=2\)

thay x=2 vào P ta có \(P=\dfrac{2^2}{2-4}=-2\)

a) \(\sqrt{x-2}+\dfrac{1}{x-5}\) có nghĩa khi:
\(\left\{{}\begin{matrix}x-2\ge0\\x-5\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\ne5\end{matrix}\right.\)

b) \(\sqrt{\left(2x-6\right)\left(7-x\right)}=\sqrt{2\left(x-3\right)\left(7-x\right)}\) có nghĩa khi:

\(\left(x-3\right)\left(7-x\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3\ge0\\7-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3\le0\\7-x\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge3\\x\le7\end{matrix}\right.\\\left\{{}\begin{matrix}x\le3\\x\ge7\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow3\le x\le7\)

c) \(\sqrt{4x^2-25}=\sqrt{\left(2x-5\right)\left(2x+5\right)}\) có nghĩa khi:

\(\left(2x-5\right)\left(2x+5\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-5\ge0\\2x+5\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-5\le0\\2x+5\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\x\ge-\dfrac{5}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{5}{2}\\x\le-\dfrac{5}{2}\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{5}{2}\\x\le-\dfrac{5}{2}\end{matrix}\right.\)

d) \(\dfrac{2}{x^2-9}-\sqrt{5-2x}=\dfrac{2}{\left(x+3\right)\left(x-3\right)}-\sqrt{5-2x}\) có nghĩa khi:

\(\left\{{}\begin{matrix}x+3\ne0\\x-3\ne0\\5-2x\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm3\\x\le\dfrac{5}{2}\end{matrix}\right.\)

e) \(\dfrac{x}{x^2-4}+\sqrt{x-2}=\dfrac{x}{\left(x+2\right)\left(x-2\right)}+\sqrt{x-2}\) có nghĩa khi:

\(\left\{{}\begin{matrix}x-2\ne0\\x+2\ne0\\x-2\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm2\\x\ge2\end{matrix}\right.\)

\(\Leftrightarrow x>2\)

A = \(\left(\dfrac{\sqrt{x}+1}{2\sqrt{x}-2}+\dfrac{3}{x-1}-\dfrac{\sqrt{x}+3}{2\sqrt{x}+2}\right)\cdot\dfrac{4x-4}{5}\) (ĐK: x \(\ge\) 0; x \(\ne\) 1)

A = \(\left(\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}+\dfrac{3}{x-1}-\dfrac{\sqrt{x}+3}{2\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{4\left(x-1\right)}{5}\)

A = \(\left(\dfrac{\left(\sqrt{x}+1\right)^2}{2\left(x-1\right)}+\dfrac{6}{2\left(x-1\right)}-\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{2\left(x-1\right)}\right)\cdot\dfrac{4\left(x-1\right)}{5}\)

A = \(\left(\dfrac{x+2\sqrt{x}+1+6-x-3\sqrt{x}+\sqrt{x}+3}{2\left(x-1\right)}\right)\cdot\dfrac{4\left(x-1\right)}{5}\)

A = \(\dfrac{10}{2\left(x-1\right)}\cdot\dfrac{4\left(x-1\right)}{5}\)

A = 4

Vậy A không phụ thuộc vào x

Chúc bn học tốt!

Ta có: \(A=\left(\dfrac{\sqrt{x}+1}{2\sqrt{x}-2}+\dfrac{3}{x-1}-\dfrac{\sqrt{x}+3}{2\sqrt{x}+2}\right)\cdot\dfrac{4x-4}{5}\)

\(=\dfrac{x+2\sqrt{x}+1+6-\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{4\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{5}\)

\(=\dfrac{x+2\sqrt{x}+7-x-2\sqrt{x}+3}{1}\cdot\dfrac{2}{5}\)

\(=10\cdot\dfrac{2}{5}=4\)