ủa lớp 8 hả bạn, mik lớp 8 mà chưa học phương trình

Nguyễn Ngọc Thạch lớp 8 kì hai nha. 

\(ĐK:x\ne-5\)

\(x^2+\frac{25x^2}{\left(x+5\right)^2}=11\)

\(\Leftrightarrow x^2-\frac{10x^2}{x+5}+\frac{25x^2}{\left(x+5\right)^2}=11-\frac{10x^2}{x+5}\)

\(\Leftrightarrow\left(x-\frac{5x}{x+5}\right)^2=11-\frac{10x^2}{x+5}\)

\(\Leftrightarrow\left(\frac{x^2}{x+5}\right)^2=11-\frac{10x^2}{x+5}\)

\(\Leftrightarrow\left(\frac{x^2}{x+5}\right)^2+\frac{10x^2}{x+5}-11=0\)

Đặt \(\frac{x^2}{x+5}=s\)thì phương trình trở thành \(s^2+10s-11=0\Leftrightarrow\left(s+11\right)\left(s-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}s=-11\\s=1\end{cases}}\)

*) \(s=-11\Rightarrow\frac{x^2}{x+5}=-11\Leftrightarrow x^2+11x+55=0\)(vô nghiệm)

*) \(s=1\Rightarrow\frac{x^2}{x+5}=1\Leftrightarrow x^2=x^2+5\)(vô nghiệm)

Vậy phương trình vô nghiệm

P/S: chưa check lại nhưng đại khái hướng làm là vậy

chtt đi

sach nang cao phat trien toan 9 tap 2 co day

Điều kiện :  \(x\ne-5\).Thêm \(-5.x.\frac{5.x}{x+5}\)vào hai vế được :

\(\left(x-\frac{5x}{x+5}\right)^2=11-\frac{25x^2}{x+5}\)

\(\Leftrightarrow\left(\frac{x^2}{x+5}\right)^2+\frac{25.x^2}{x+5}-11=0\)

Đạt  \(\frac{x^2}{x+5}=y\),ta được \(y^2+5y-11=0\)(1)

Mà \(y^2+5y-11\ge0\forall y\)

Nên phương trình (1)  vô nghiệm 

\(\Rightarrow x^2+\frac{25x^2}{\left(x+5\right)^2}=11\)vô nghiệm

Vậy không có giá tri nào của x thỏa mãn phương trình trên

\(BPT\Leftrightarrow x\sqrt[3]{25x\left(2x^2+9\right)}\le4x^2+3\\ \Leftrightarrow\sqrt[3]{25x^4\left(2x^2+9\right)}\le4x^2+3\left(1\right)\)

Áp dụng BĐT cosi:

\(\sqrt[3]{5x^2\cdot5x^2\left(2x^2+9\right)}\le\dfrac{5x^2+5x^2+2x^2+9}{3}=\dfrac{12x^2+9}{3}=4x^2+3\)

Vậy \(\left(1\right)\) luôn đúng

Dấu \("="\Leftrightarrow5x^2=2x^2+9\Leftrightarrow x^2=3\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\)

a: =>3x-9+5+10x=90

=>13x-4=90

=>13x=94

hay x=94/13

b: \(\Leftrightarrow2x-4-x-1=3x-11\)

=>3x-11=x-5

=>2x=6

hay x=3(nhận)

\(\dfrac{x^2\left(x+5\right)^2+25x^2-11\left(x+5\right)^2}{\left(x+5\right)^2}=\dfrac{x^4+10x^3+50x^2-11x^2-11.2.5x-11.5.5}{\left(x+5\right)^2}\)

\(\dfrac{\left[x^4-x^3-5x^2\right]+\left[11x^3-11x^2-11.5x\right]+55x^2-55x-5.55}{\left(x+5\right)^2}\)

\(\dfrac{x^2\left(x^2-x-5\right)+11x\left(x^2-x-5\right)+55\left(x^2-x-5\right)}{\left(x+5\right)^2}=\dfrac{\left(x^2-x-5\right)\left(x^2+11x+55\right)}{\left(x+5\right)^2}\)

\(\left[{}\begin{matrix}x^2-x-5=0\left(1\right)\\x^2+11x+55=0\left(2\right)\end{matrix}\right.\)

(2) vô nghiệm

(1)\(\Leftrightarrow\) \(\left[x^2-\dfrac{1}{2}x+\dfrac{1}{4}\right]=\dfrac{21}{4}\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{21}{4}\)

\(\left[{}\begin{matrix}x=\dfrac{1-\sqrt{21}}{2}\\x=\dfrac{1+\sqrt{21}}{2}\end{matrix}\right.\)

\(x^2+\dfrac{25x^2}{\left(x+5\right)^2}=11\) (ĐKXĐ: \(x\ne-5\))

\(\Leftrightarrow\dfrac{x^2\left(x+5\right)^2+25x^2}{\left(x+5\right)^2}=\dfrac{11\left(x+5\right)^2}{\left(x+5\right)^2}\)

\(\Rightarrow x^4+10x^3+25x^2+25x^2-11x^2-110x-275=0\\ \Leftrightarrow x^4+10x^3+39x^2-110x-275=0\)

mình ko biết phân tích sao nữa

Ta có: \(x^{2^{ }}+\dfrac{25x^2}{\left(x+5\right)^2}=11\)

ĐKXĐ: x\(\ne\)-5

<=> \(x^2-2x\dfrac{5x}{x+5}+\dfrac{25x^2}{\left(x+5\right)^2}+2x\dfrac{5x}{x+5}=11\)

<=> \(\left(x-\dfrac{5x}{x+5}\right)^2+10\dfrac{x^2}{x+5}=11\)

<=> \(\left(\dfrac{x^2}{x+5}\right)^2+10\dfrac{x^2}{x+5}=11\)

Gọi \(\dfrac{x^2}{x+5}=t\)

=> t²+10t = 11

<=> t²+10t-11=0

<=> (t-1)(t+1)+10(t-1)=0

<=>(t-1)(t+11)=0

=>\(\left[{}\begin{matrix}t-1=0\\t+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-11\end{matrix}\right.\)

hay: \(\left[{}\begin{matrix}\dfrac{x^2}{x+5}=1\\\dfrac{x^2}{x+5}=-11\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{21}}{2}\\x=\dfrac{1+\sqrt{21}}{2}\end{matrix}\right.\)

\(ĐKXĐ:\left\{{}\begin{matrix}x\ne-1\\x\ne2\end{matrix}\right.\)

\(\dfrac{2}{x+1}-\dfrac{1}{x-2}=\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\\ \Leftrightarrow\dfrac{2\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}-\dfrac{x+1}{\left(x+1\right)\left(x-2\right)}=\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\\ \Leftrightarrow2\left(x-2\right)-\left(x+1\right)=3x-11\\ \Leftrightarrow2x-4-x-1-3x+11=0\\ \Leftrightarrow-2x+6=0\\ \Leftrightarrow-2x=-6\\ \Leftrightarrow x=3\left(tm\right)\)

\(\Leftrightarrow\dfrac{2\left(x-2\right)-\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\)

\(\Leftrightarrow2x-4-x-1=3x-11\left(khử\cdot mẫu\right)\)

\(\Leftrightarrow2x-x-3x=-11+4+1\)

\(\Leftrightarrow-2x=-6\)

\(\Leftrightarrow x=3\)

Vậy \(S=\left\{3\right\}\)