chứng minh \(\dfrac{x-3}{x}-1>2x\)
1.chứng minh \(\dfrac{6x^3-x^6}{x^4-2x^2+4}< 3\) với mọi x ∈ R
2.chứng minh \(\dfrac{x^4-4x^2+8}{2x-x^2}>4\) với mọi x ∈ (0;2)
Chứng minh:
1.\(\dfrac{\cot^2x-\sin^2x}{\cot^2x-\tan^2x}=\sin^2x\cdot\cos^2x\)
2.\(\dfrac{1-\sin x}{\cos x}-\dfrac{\cos x}{1+\sin x}=0\)
3.\(\dfrac{\tan x}{\sin x}-\dfrac{\sin x}{\cot x}=\cos x\)
4.\(\dfrac{\tan x}{1-\tan^2x}\cdot\dfrac{\cot^2x-1}{\cot x}=1\)
5.\(\dfrac{1+\sin^2x}{1-\sin^2x}=1+2\tan^2x\)
Câu 1 đề sai, chắc chắn 1 trong 2 cái \(cot^2x\) phải có 1 cái là \(cos^2x\)
2.
\(\dfrac{1-sinx}{cosx}-\dfrac{cosx}{1+sinx}=\dfrac{\left(1-sinx\right)\left(1+sinx\right)-cos^2x}{cosx\left(1+sinx\right)}=\dfrac{1-sin^2x-cos^2x}{cosx\left(1+sinx\right)}\)
\(=\dfrac{1-\left(sin^2x+cos^2x\right)}{cosx\left(1+sinx\right)}=\dfrac{1-1}{cosx\left(1+sinx\right)}=0\)
3.
\(\dfrac{tanx}{sinx}-\dfrac{sinx}{cotx}=\dfrac{tanx.cotx-sin^2x}{sinx.cotx}=\dfrac{1-sin^2x}{sinx.\dfrac{cosx}{sinx}}=\dfrac{cos^2x}{cosx}=cosx\)
4.
\(\dfrac{tanx}{1-tan^2x}.\dfrac{cot^2x-1}{cotx}=\dfrac{tanx}{1-tan^2x}.\dfrac{\dfrac{1}{tan^2x}-1}{\dfrac{1}{tanx}}=\dfrac{tanx}{1-tan^2x}.\dfrac{1-tan^2x}{tanx}=1\)
5.
\(\dfrac{1+sin^2x}{1-sin^2x}=\dfrac{1+sin^2x}{cos^2x}=\dfrac{1}{cos^2x}+tan^2x=\dfrac{sin^2x+cos^2x}{cos^2x}+tan^2x\)
\(=tan^2x+1+tan^2x=1+2tan^2x\)
a, Cho x, y, z > 0 \(\in[0,1]\). Chứng minh:
\(\dfrac{x}{yz+1}+\dfrac{y}{xz+1}+\dfrac{z}{xy+1}< 2\)
b, x, y, z > 0 : xyz = 1. Chứng minh:
\(\dfrac{1}{x^2+2y+3}+\dfrac{1}{y^2+2z^2+3}+\dfrac{1}{z^2+2x^2+3}\le2\)
Chứng minh đẳng thức:
a, \(\left(\dfrac{3}{2x-y}-\dfrac{2}{2x+y}-\dfrac{1}{2x-5y}\right).\dfrac{4x^2-y^2}{y^2}=\dfrac{-24}{2x-5y}\)
b, \(\dfrac{x^2-x+1}{x^2+x}.\dfrac{x+1}{3x-2}.\dfrac{9x-6}{x^2-x+1}=\dfrac{3}{x}\)
Lời giải
a)
\(\left(\frac{3}{2x-y}-\frac{2}{2x+y}-\frac{1}{2x-5y}\right).\frac{4x^2-y^2}{y^2}\)
\(=\frac{3(4x^2-y^2)}{(2x-y)y^2}-\frac{2(4x^2-y^2)}{(2x+y)y^2}-\frac{4x^2-y^2}{(2x-5y)y^2}\)
\(=\frac{3(2x-y)(2x+y)}{(2x-y)y^2}-\frac{2(2x-y)(2x+y)}{(2x+y)y^2}-\frac{4x^2-y^2}{(2x-5y)y^2}\)
\(=\frac{3(2x+y)-2(2x-y)}{y^2}-\frac{4x^2}{(2x-5y)y^2}+\frac{1}{2x-5y}\)
\(=\frac{2x+5y}{y^2}-\frac{4x^2}{(2x-5y)y^2}+\frac{1}{2x-5y}\)
\(=\frac{(2x+5y)(2x-5y)-4x^2}{(2x-5y)y^2}+\frac{1}{2x-5y}\)
\(=\frac{4x^2-25y^2-4x^2}{(2x-5y)y^2}+\frac{1}{2x-5y}=\frac{-25}{2x-5y}+\frac{1}{2x-5y}=\frac{-24}{2x-5y}\)
Ta có đpcm.
b)
\(\frac{x^2-x+1}{x^2+x}.\frac{x+1}{3x-2}.\frac{9x-6}{x^2-x+1}\)
\(=\frac{(x^2-x+1)(x+1).3(3x-2)}{x(x+1)(3x-2)(x^2-x+1)}\)
\(=\frac{3}{x}\) (đpcm)
A=(\(\dfrac{2}{x+1}\)-\(\dfrac{1}{x-1}\)+\(\dfrac{5}{x^2-1}\)):\(\dfrac{2x+1}{x-1}\)
Chứng minh A=\(\dfrac{x+2}{2x-1}\)
\(A=\left(\dfrac{2}{x+1}-\dfrac{1}{x-1}+\dfrac{5}{x^2-1}\right):\dfrac{2x+1}{x-1}\\ =\left(\dfrac{2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x+1}{\left(x+1\right)\left(x-1\right)}+\dfrac{5}{\left(x+1\right)\left(x-1\right)}\right).\dfrac{x-1}{2x+1}\\ =\dfrac{2x-2-x-1+5}{\left(x+1\right)\left(x-1\right)}.\dfrac{x-1}{2x+1}\\ =\dfrac{x+2}{\left(x+1\right)\left(2x+1\right)}\)
Đề sai r bn
chứng minh rằng:
\(\dfrac{x+2}{x-1}.\left(\dfrac{x^3}{2x+2}+1\right)-\dfrac{8x+7}{2x^2-2}>0\)
\(=\dfrac{x^3\left(x+2\right)}{2\left(x-1\right)\left(x+1\right)}+\dfrac{x+2}{x-1}-\dfrac{8x+7}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^4+2x^3+2\left(x+1\right)\left(x+2\right)-8x-7}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^4+2x^3+2x^2+6x+4-8x-7}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^4+2x^3+2x^2-2x-3}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x-1\right)\left(x+1\right)\left(x^2+2x+3\right)}{2\left(x-1\right)\left(x+1\right)}=\dfrac{x^2+2x+3}{2}>0\)
A=(\(\dfrac{2}{x+1}\)-\(\dfrac{1}{x-1}\)+\(\dfrac{5}{x^2-1}\)):\(\dfrac{2x-1}{x^2-1}\)
Chứng minh A=\(\dfrac{x+2}{2x-1}\)
\(A=\left(\dfrac{2}{x+1}-\dfrac{1}{x-1}+\dfrac{5}{x^2-1}\right):\dfrac{2x+1}{x^2-1}\\ =\left(\dfrac{2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x+1}{\left(x+1\right)\left(x-1\right)}+\dfrac{5}{\left(x+1\right)\left(x-1\right)}\right).\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\\ =\dfrac{2x-2-x-1+5}{\left(x+1\right)\left(x-1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\\ =\dfrac{x+2}{2x+1}\)
BT1:
a) 2x-1=0 ; b) 3x-2=5+x ; c) 2(x-3)-4=3(1+x)-5x ; d) \(\dfrac{x+1}{2}\)- \(\dfrac{2x}{3}\)=1 ; e) x(x-2)+3(x-2)=0 ; f) \(\dfrac{x+1}{x-1}\)+ \(\dfrac{3}{x}\)= \(\dfrac{x^2+2}{x^2-x}\)
BT2:
a) Cho a>b, chứng minh rằng 2a+1>2b-3
b) Tìm x để giá trị của biểu thức 3x-1 ≤ giá trị biểu thức x+2
c) Giải các bất phương trình sau và biểu diễn tập nghiệm trên trục số (mng giúp mình giải phương trình thôi nha)
2x+3>0 ; 3x+1<x-4 ; 2(x+1)+3≥ 3(5-x) ; \(\dfrac{x}{3}\)-\(\dfrac{x+1}{5}\)>1
BT3: Giải bài toán bằng cách lập phương trình
1 ô tô đi từ A đến B với vận tốc 50km/h. Đến B, ô tô nghỉ lại 1h, sau đó quay trở về A với vận tốc 60km/h. Tổng thời gian đi và về(gồm thời gian nghỉ lại) là 6h30p. Tính quãng đường AB?
Mng giúp mình với mai mình kiểm tra rồi ạ, mình cảm ơn
cho 3 số dương x,y,z thỏa mãn \(\dfrac{1}{x}\)+\(\dfrac{1}{y}\)+\(\dfrac{1}{z}\)=4. Chứng minh:
\(\dfrac{1}{2x+y+z}\)+\(\dfrac{1}{x+2y+z}\)+\(\dfrac{1}{x+y+2z}\)≤1
Bổ đề:\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\Leftrightarrow\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
Ta có:\(\dfrac{1}{2x+y+z}\le\dfrac{1}{4}\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}\right)\le\dfrac{1}{4}.\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{x}+\dfrac{1}{z}\right)\)
Tương tự ta có:\(\dfrac{1}{2y+z+x}\le\dfrac{1}{4}.\dfrac{1}{4}\left(\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{y}+\dfrac{1}{x}\right)\)
\(\dfrac{1}{2z+x+y}\le\dfrac{1}{4}.\dfrac{1}{4}\left(\dfrac{1}{z}+\dfrac{1}{x}+\dfrac{1}{z}+\dfrac{1}{y}\right)\)
Cộng vế với vế ta có:
\(\dfrac{1}{2x+y+z}+\dfrac{1}{2y+z+x}+\dfrac{1}{2z+x+y}\le\dfrac{1}{16}\left[4\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\right]=\dfrac{1}{16}.4.4=1\)
Dấu "=" xảy ra ⇔ \(x=y=z=\dfrac{3}{4}\)