Giúp tôi giải toán và làm văn

x\(\ge\frac{3}{2}\)nha

Bổ sung : ĐKXĐ: x\(\ne1\)

ĐKXĐ: \(x\le1\)hoặc \(x\ge\frac{3}{2}\)

Bình phương 2 vế ta được:

\(\frac{2x-3}{x-1}=2\)

<=> 2x-3=2x-2

<=> 2x-3-2x+2=0

<=> -1=0 ( vô lí)

Vậy không tìm được x thỏa mãn yêu cầu đề bài

haizz .Lại nhìn thiếu rồi :(

\(x^2\left(x-2\right)^2=0\)

\(\Rightarrow\orbr{\begin{cases}x^2=0\\\left(x-2\right)^2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}}\)

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

\(\Leftrightarrow x^2+2=2\sqrt{x^3+1^3}\)

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

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

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

\(\Leftrightarrow x^4+4x^2+4=4x^3-4x^2+4x+4x^2-4x+4\)

\(\Leftrightarrow x^4+4x^2=4x^3\)

\(\Leftrightarrow x^4+4x^2-4x^3=0\)

\(\Leftrightarrow x^2\left(x^2+4-4x\right)=0\)

\(\Leftrightarrow x^2\left[x^2-2\left(x\right)\left(2\right)+2^2\right]=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

Vậy nghiệm phương trình là: {0; 2}

\(x^2+2=2\sqrt{x^3+1}\)

\(\Rightarrow x^4+4x^2+4=4\left(x^3+1\right)\)

\(\Rightarrow x^4+4x^2+4-4x^3-4=0\)

\(\Rightarrow x^4+4x^2-4x^3=0\)

\(\Rightarrow x^2\left(x^2-4x+4\right)=0\)

\(\Rightarrow\left(x-2\right)^2=0\)

\(\Rightarrow x=2\) 

khi tinh sai

1+1=3

khi bạn làm sai

1+1=3 khi làm sai.

Có đúng ko?

ko có x để B= 1/2

\(\Leftrightarrow1+\sqrt{x^2+7x+10}=\sqrt{x+5}+\sqrt{x+2}\)

\(\Leftrightarrow\sqrt{x^2+7x+10}-2-\sqrt{x+5}+2-\sqrt{x+2}+1=0\)

\(\Leftrightarrow\frac{\left(x+1\right)\left(x+6\right)}{\sqrt{x^2+7x+10}+2}+\frac{x+1}{2+\sqrt{x+5}}+\frac{x+1}{1+\sqrt{x+2}}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{x+6}{\sqrt{x^2+7x+10}+2}+\frac{1}{2+\sqrt{x+5}}+\frac{1}{1+\sqrt{x+2}}\right)=0\)

Giải nốt nhá ^.^

\(\left(\sqrt{x+5}-\sqrt{x+2}\right)\left(1+\sqrt{x^2+7x+10}\right)=3.\)

\(\Rightarrow\left(\sqrt{x+5}-\sqrt{x+2}\right)\left(1+\sqrt{\left(x+2\right)\left(x+5\right)}\right)=3\)

Đặt : \(\sqrt{x+5}=a\Rightarrow x+5=a^2\)

\(\sqrt{x+2}=b\Rightarrow x+2=b^2\)\(\left(đk:a,b\ge0\right)\)

\(\Rightarrow a^2-b^2=x+5-x-2=3\left(1\right)\)

Mà theo phương trình, ta có :

\(\left(a-b\right)\left(1+ab\right)=3\)

\(\Rightarrow a+a^2b-b-ab^2=3\)\(\left(2\right)\)

Tự giải hệ 

Trả lời

H2O (nước) + NaOH (natri hidroxit)

= [NA(H2O)4] (Tetraaquasodium ion)

Bài làm

H₂O + NaOH ---------> OH + [ Na ( H₂O )₄ ]

~ 1000% đúng ~
# Học tốt #

Trả lời 

4H₂O + NaOH------> OH^- + [ Na ( H₂O )₄ ] ⁺

Cũng không chắc nx

Study well 

\(\sqrt{2}A=\sqrt{4+2\sqrt{3}}-\sqrt{12-6\sqrt{3}}+\frac{2}{\sqrt{3}+1}\)

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

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

\(A=\sqrt{6}-\sqrt{2}+\frac{\sqrt{2}}{1+\sqrt{3}}=\frac{3\sqrt{6}-3\sqrt{2}}{2}\)

\(đkxđ\Leftrightarrow\orbr{\begin{cases}x\ge0\\x\ne9\end{cases}}\)

\(P=\left(\frac{\sqrt{x}}{3-\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right).\)

\(=\left(\frac{\sqrt{x}}{3-\sqrt{x}}+\frac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\)\(\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)

\(=\frac{\sqrt{x}\left(3+\sqrt{x}\right)+x+9}{\left(3-\sqrt{x}\right)}:\)\(\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\frac{3\sqrt{x}+x+x+9}{3-\sqrt{x}}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}\)

\(=-\frac{\sqrt{x}\left(2x+3\sqrt{x}+9\right)}{2\sqrt{x}+4}\)?

h.vn là gì bạn

Toán lp 9 

Thì mk khuyên bn sang h.vn để đc giải đáp tốt hơn

Nhìn thấy thì tiếc gì nhỉ =)

\(P=\frac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+1-\frac{2x+\sqrt{x}}{\sqrt{x}}\left(đkxđ\Leftrightarrow x\ge0\right).\)

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

\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}-\left(2\sqrt{x}+1\right)\)

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

\(P=x-\sqrt{x}-1=\sqrt{x}^2-2.\sqrt{x}.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}-1\)

\(=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{5}{4}\)

\(\Rightarrow P_{min}=-\frac{5}{4}\Leftrightarrow\left(\sqrt{x}-\frac{1}{2}\right)^2=0\)

\(\Rightarrow\sqrt{x}=\frac{1}{2}\Rightarrow x=\frac{1}{4}\)